
flass Q b ft ? 
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CCEUUGHT DEPOSEt 



THE MARSH AND ASHTON MATHEMATICAL SERIES. 

BY 

WALTER R. MARSH, 

HEAD MASTER PINGRY SCHOOL. ELIZABETH. N.J. 

AND 

CHARLES H. ASHTON, 
INSTRUCTOR IN MATHEMATICS, HARVARD UNIVERSITY. 



The series will include text-books in 

Elementary Algebra. 

College Algebra. 

Plane and Solid Geometry. 

Plane and Spherical Trigonometry. 

Plane and Solid Analytic Geometry. 



PLANE AND SOLID 
ANALYTIC GEOMETRY 



AN ELEMENTARY TEXT-BOOK 



I 
CHARLES H. ASHTON, A.M. 

INSTRUCTOR IN MATHEMATICS IN HARVARD UNIVERSITY 



NEW YORK 
CHARLES SCRIBNER'S SONS 
1901 
U - 



90203 



Library of Congress 

Two Copies Received 
DEC 18 1900 

Copyright «*ry 



No 



SECOND COPY 

OflKvwid to 
ORDER OIVtSION 
JAN 4 1901 



COPYRIGHT, 1900, BY 
CHARLES SCRIBNER'S SONS 



Norfooorj press 

J. S. dishing & Co. -Berwick & Smith 
Norwood Mass. U.S. A 



PREFACE 

The present work is intended as a text-book for class- 
room use, and not as ari" exhaustive treatise on the sub- 
ject. This object has been kept constantly in mind in 
writing the book, and every, subject has been treated from 
this point of view. A large part of the book was mimeo- 
graphed and tested by use for several years in the author's 
classes in Harvard University. 

The author has tried to meet the needs of a class which 
occupies from sixty to seventy recitation hours upon the 
subject, and it is thought that the book ought to be com- 
pleted by the average class in that time. Necessarily 
some subjects which usually find a place in books on 
Analytic Geometry have been omitted ; but it is thought 
that nothing has been omitted which has an important 
bearing on future mathematical study. 

The conies have been treated from their ratio defini- 
tion, and much space and time have been gained by not 
repeating proofs which are identical, or very similar, for 
the three forms of the conic. Analytic methods are used 
throughout the book, and the author has attempted to 
give proofs which are concise and easily understood by 
the average student, but, at the same time, mathemati- 
cally rigorous. In this connection he would call atten- 
tion to the proofs in oblique coordinates (Arts. 12, 28), 
which are usually given without reference to the direc- 
tions of the lines, and, therefore, do not hold if the 
positions of the points are changed. 



VI PREFACE 

Numerous problems, which have been selected with 
great care, have been inserted after nearly every article. 
In the early part of the book these are mainly numerical, 
but later the student has been asked to prove a large 
number of theorems. A considerable number of theorems 
which are usually proved in the text are here inserted 
as problems, and in many places the student has been 
asked to derive formulas for two of the conies, after the 
corresponding formulas for one of the conies have been 
obtained. It is only by solving such problems that the 
student can acquire any real grasp of the subject. 

The attention of the teacher is also called to the two 
chapters on loci (Chaps. VIII and XIV), in which a large 
number of problems are given and the methods of solving 
them discussed ; to the treatment of poles and polars by 
the aid of harmonic division (Chap. XII); and to the 
system of polar coordinates used in the Solid Geometry. 

The author desires to acknowledge gratefully the assist- 
ance of Mr. B. E. Carter of the Massachusetts Institute 
of Technology, who has read with great care both manu- 
script and proof ; of Mr. E. V. Huntington, who made 
many valuable suggestions during the early stages of 
the work ; and of Mr. W. R. Marsh, his colleague in the 
preparation of the series, of which this volume is the first 
to appear. 



Cambridge, 

November, 1900. 



CONTENTS 

PAKT I 
PLANE ANALYTIC GEOMETRY 

CHAPTER I 
Introduction 

ART. PAGE 

1. Directed lines 1 

2. Addition of directed lines 1 

3. Directed angles 2 

4. Addition of directed angles 2 

5. Measurement of lines and angles ...... 3 

6. Angles between two lines -1 

7. Law of sines and law of cosines .;.... 5 

8. The quadratic equation 6 

CHAPTER II 
The Point 

9. Cartesian coordinate systems . 7 

10. Notation 11 

11. Distance between two points in rectangular coordinates . 11 

12. Distance between two points in oblique coordinates . . 13 

13. Points dividing a line in a given ratio 14 

14. Harmonic division 16 

CHAPTER III 
Loci 

15. Equation of a locus 20 

16. Locus of an equation 24 

17. Plotting the locus of an equation 25 



viii CONTENTS 

ART. PA<;i 

18. Symmetry 27 

19. Intercepts 31 

20. Intersection of two curves 31 

21. Locus of u -f kv = 0, and uv = 31 



CHAPTER IV 
The Straight Line 

22. Introduction 34 

23. Line through two points 34 

24. Line determined by its intercepts 36 

25. Oblique coordinates . . . . . . .30 

26. Line determined by a point and its direction . . .37 

27. Line determined by its slope and its intercept on the Y-axis 38 

28. Oblique coordinates 39 

29. General equation of the first degree 40 

30. Two equations representing the same line cannot differ 

except by a constant factor 42 

31. The angle which one line makes with another ... 44 

32. Perpendicular and parallel lines 46 

33. Line making a given angle with a given line . . .48 

34. Xormal form of the equation of a straight line ... 49 

35. Reduction of the general equation to the normal form . 51 

36. Distance of a point from a line 53 

37. Oblique coordinates .55 

38. Bisector of the angle between two lines . . . .56 

39. Lines through the intersection of two given lines . . .57 

40. Area of a triangle 59 

CHAPTER V 
Polar Coordinates 

41. Introduction ..63 

42. Equation of a locus .64 

43. Plotting in polar coordinates 65 

44. Natural values of the sines, cosines, tangents, and cotangents 66 



CONTENTS ix 

CHAPTER VI 
Transformation of Coordinates 

ART. PAGE 

45. Introduction . . . . . . . ... .68 

46. Transformation to axes parallel to the original axes . . 68 

47. Transformation from one set of rectangular axes to another, 

having the same origin and making an angle 6 with the 
first set 70 

48. Transformation in which both the position of the origin 

and the direction of the axes are changed .... 71 

49. Transformation from any Cartesian system to any other 

Cartesian system, having the same origin .... 72 

50. Degree of an equation not changed by transformation of 

coordinates 72 

51. Transformation from rectangular to polar coordinates . . 73 



CHAPTER VII 
The Circle 

52. Equation 75 

53. General form of the equation 76 

54. Circle through three points 77 

55. Tangent 79 

56. Normal 81 

57. Tangents from an exterior point 82 

58. Tangent in terms of its slope 84 

59. Chord of contact 85 



CHAPTER VIII 
Loci 

60. Problems 88 

61. Problems . . . 91 



CONTENTS 



CHAPTER IX 



Conic Sections 

ART. PACtE 

62. Definition and equation 100 

63. Parabola . . 101 

64. Central conies 103 

(55. Ellipse . 106 

66. Hyperbola 110 

67. Asymptotes Ill 

68. Conjugate hyperbolas 116 

69. Equilateral or rectangular hyperbola 117 

70. Focal radii of a central conic 118 

71. Mechanical construction of the conies . . . . . 120 

72. Auxiliary circles 121 

73. General equation of conies when axes are parallel to the 

coordinate axes 123 



CHAPTER X 
Tangents 

74. Equations of tangents 126 

75. Normals 128 

76. Subtangents and subnormals 129 

77. Slope form of the equations of tangents .... 131 

78. Theorems concerning tangents and normals .... 133 



CHAPTER XI 
Diameters 



79. Equations of diameters 142 

80. Conjugate diameters 144 

81. Equation of conjugate diameter 147 

82. Theorems concerning diameters 148 



CONTENTS 



CHAPTER XII 
Poles and Polars. 

ART. PAGE 

83. Harmonic division 154 

84. Polar of a point . .155 

85. Position of the polar 157 

86. Theorems concerning poles and polars . 159 



CHAPTER XIII 
General Equation of the Second Degree 

87. Introduction 166 

88. Two straight lines ' . . . .166 

B* - 4 A C =£ 0. 

89. Removal of the terms of the first degree .... 168 

90. Removal of the term in xy 169 

91. Determination of the coefficients A', C, and F' . . . 170 

92. Nature of the locus 172 

B 2 -±AC=0. 

93. Removal of the term in xy 175 

94. Removal of the term in y 177 

95. Nature of the locus 177 

96. Second method of reducing the general equation to a simple 

form, when B - 4 A C = ....... 178 

97. Summary 182 

98. General equation in oblique coordinates . . . .183 

99. Conic through five points . 183 

CHAPTER XIV 

Problems in Loci , 185 



CONTENTS 



PAKT II 
ANALYTIC GEOMETRY OF SPACE 

CHAPTER 
Coordinate Systems. The Point 

ART. PAGE 

1. Introductory 195 

2. Rectangular coordinates 196 

3. Distance between two points 197 

4. To divide a line in any given ratio 198 

5. Projection of a given line on a given axis .... 199 

6. Polar coordinates . . . 200 

7. Spherical coordinates 203 

8. Angle between two lines 204 

9. Transformation of coordinates. Parallel axes . . . 206 

10. Transformation of coordinates from ' one set of rectangular 

axes to another which has the same origin . . . 206 

CHAPTER II 
Loci 

11. Equation of a locus 208 

12. Cylindrical surfaces 209 

13. Surfaces of revolution 209 

14. Locus of an equation 211 

CHAPTER III 
The Plane 

15. Normal form of the equation of a plane .... 215 

16. Reduction of the general equation Ax + By + Cz + D = to 

the normal form 216 

17. Equation of a plane in terms of its intercepts . . . 217 

18. Distance of a point from a plane 217 



CONTENTS Xlll 

ART. PAGE 

19. The angle between two planes 219 

20. Perpendicular and parallel planes 220 

21. Equation of a plane satisfying three conditions . . . 220 



CHAPTER IV 
The Straight Line 

22. Equations 223 

23. The equations of a line in terms of its direction cosines and 

the coordinates of a point through which it passes . . 225 

24. Given the equations of a line, to find its direction cosines . 226 

25. Equations of a line through two points 228 

CHAPTER V 
Quadric Surfaces 

26. The sphere 230 

27. Conicoids 232 

28. The ellipsoid 233 

29. The unparted hyperboloid 235 

30. The biparted hyperboloid 238 

31. The cone 240 

32. Asymptotic cones 241 

33. The paraboloids 242 

34. Ruled surfaces 245 

35. Tangent planes 247 

36. Normals 249 

37. Diametral planes 250 

38. Polar plane 252 

Answers 257 



PART I 

PLANE ANALYTIC GEOMETRY 

CHAPTER I 

INTRODUCTION 

1. Directed lines. — If a point moves from A to B in 
a straight line, Ave shall say that it generates the line 
A B ; if it moves from B to A, it generates the line BA. 
In our study of Geometry, AB ^ ^ - c 



and BA meant the same thing, — | ^ B 

the line joining A and B with- c j 7^ b 

out regard to its direction. But FlG - \ 

we shall now find it convenient to distinguish betjupen AB 
and BA as if they were separate lines. The positi^i from 
which the generating point starts is called the initial point 
of the line ; the point where it stops, the terminal point. 

2. Addition of directed lines. — If a point moves in a 
straight line from A to B (on any one of the lines in 
Fig. 1) and then moves in that line, or in that line pro- 
duced, to (7, the position which it finally reaches is evi- 
dently the same as if, starting from A, it had moved 
along the single line AC. The line AC is called the sum 
of the lines AB and BC; that is, AB + BC= AC. Evi- 
dently AB + BA = AA = 0, and hence AB = - BA. 

1 



2 ANALYTIC GEOMETRY [Ch. I, §§ 3, 4 

3. Directed angles. — If a line starts from the position 
OA and rotates in a fixed plane about the point into 
the position OB, it is said to generate the angle A OB. 
If it rotates from OB to OA, it generates the angle BOA. 

We shall find it convenient to 

S~ X^ distinguish between the angles 

/ ^ — ^ /^ A OB and BOA as if they were 

/ / /\ 2 \ separate angles. The position 

\4 \s o^- — -| \a from which the moving line 

\ \ / I starts is called the initial side 

N* S / of the angle ; the position where 

N v v W S it stops, the terminal side. 

There is no limit to the pos- 
sible amount of rotation of the 
moving line ; after performing a complete revolution in 
either direction, it may continue to rotate as many 
times as we please, generating angles of any magnitude 
in either direction. Angles which are not equal, but 
have the same initial and terminal sides (1 and 3, or 2 
and 4, Fig. 2) are called congruent angles. 

In reading the angle AOB, it is not possible to dis- 
tinguish between the various congruent angles which 
have OA and OB for their initial and terminal lines, but 
we shall understand that the smallest of the congruent- 
angles is meant, unless another angle is indicated by an 
arrow in the figure. 

4. Addition of directed angles. — If the moving line 
starts from OA (in any one of these figures) and rotates 
first through the angle A OB, and then through the angle 
BOC, it is evident that the position 00. which the line 



Ch. I, § 5] 



INTRODUCTION 



3 



finally reaches, is the same as if, starting from OA, it 
had rotated through the single angle A OC. The angle 



— \B 






Fig. 3. 

A OC is called the sum of the angles A OB and BOC; 
that is, ZAOB+ZBOC=ZAOC. Evidently ZAOB + 
ZBOA = 0, and hence ZAOB = - ZBOA. 

5. Measurement of lines and angles. — The length of 
a line, or the magnitude of an angle, may be represented 
by a number, by the familiar process of measurement. 
That is, the number of times which the given line or 
angle contains an arbitrarily chosen unit may be used 
to represent the length of the line or the magnitude of 
the angle. But we have seen that it is necessary to 
distinguish between the lines AB and BA, and that it 
has followed from our definition of addition of lines that 
AB = — BA. Hence, if the line AB is represented by a 
positive number, the line BA will be represented by the 
same number with a negative sign. In like manner, if 
the angle A OB is represented by a positive number, the 
angle BOA will be represented by the same number with 
a negative sign. It follows, therefore, that opposite direc- 
tions are indicated by opposite signs ; that is, if the length 
of a line or the magnitude of an angle, generated in one direc- 
tion, is represented by a positive number, then the length of 
a line or the magnitude of an angle generated in the opposite 



ANALYTIC GEOMETRY 



[Ch. I, § 6 



direction, is represented by a negative number. Either of 
two opposite directions may be chosen as the positive 
direction ; then the other must be taken as the negative. 

As all our work will be concerned with the algebraic 
number rather than the geometric line which it represents, 
it will not be necessary to distinguish between the line AB 
and the number which represents its length. We shall 
let AB stand for the number which represents the length of 
the line from A to B. It is easily shown that the length 
of the sum of two or more lines that run in the same or 
in opposite directions is the algebraic sum of the lengths 
of the separate lines. Hence it is still true that 
AB+BC= AC, when AB, BO, and AC stand for the 
lengths of the lines AB, BC, and AC. Since these are 
now algebraic numbers, it follows that AB — AC — BC. 

In like manner A OB will be used to represent the 
magnitude of the angle instead of the angle itself. With 
this meaning it will still be true that 

Z AOB + Z BOC = Z AOC. 



Also, 



Fig. 4. 



Z AOB= Z AOC- ZBOC. 

6. Angles between two lines. — 
When two lines intersect at a point, 
they form several angles at that point. 
To avoid ambiguity, if the lines are 
directed lines, we shall define the 
angle between them as the angle from 
the positive direction of the first line to 
the positive direction of the second line, 
the smallest of the congruent angles 
being chosen. 




Ch. I, § 7] INTRODUCTION 5 

We shall adopt the following notation : Denoting the 
intersecting lines by single letters, as a and 6, the symbol 
ab shall indicate the angle from the positive direction of 
the line a to the positive direction of the line b, to be 
read, "the angle from a to 6." 

It will sometimes be inconvenient to choose either direc- 
tion of an unlimited line as positive. (As when the line 
is given by its equation.) We shall then define the angle 
which one line makes with another as the angle formed in 
going from the second to the first in the positive direc- 
tion of rotation. 

It is customary to call the angle from a to b positive if 
its rotation is opposite to that of the hands of a clock ; nega- 
tive if in the same direction as the hands. 

7. Law of sines and law of cosines. — The two laws 
concerning the sines and the cosines of the angles of a 
triangle are often stated in trigonometry without regard 
to the direction of the sides of 
the triangle. But for our work 
these must be stated in a more 
accurate form. Let the posi- 
tive direction of each side of 
the triangle ABO be fixed. It 
can be easily shown that these 

two laws take the following form, when the directions of 
the lines and angles are considered. 

T £ . AB sin ab 

Law or sines : == — • 

BC sin be 

Law of cosines : 

(AB) 2 = (BO)* + (OA) 2 + 2(BO)(OA) cos ab. 




6 ANALYTIC GEOMETRY [Ch. I, § 8 

8. The quadratic equation. — We shall have occasion to 
use a few theorems in quadratic equations which it seems 
advisable to reproduce here. 

Any quadratic equation may be written in the form 



ax 2 + bx + e = 0. 






The two roots of this equation are 






_ h + V6 2 - 4-<w , - b - 


-V6 2 - 


- 4 ae 


Jy-I , ClllKX ,t(j 


2 a 




By addition, x 1 -f x 2 = 






By multiplication, x-^x 2 = — 







The sum and the product of the roots can therefore be 
found directly from the equation without solving. 

The character of the roots depends on the quantity 
under the radical, b 2 — 4 ae. 

If b 2 — 4 ac > 0, the roots are real and unequal, 

if b 2 — 4 ae = 0, the roots are real and equal, 

if b 2 — 4 ae < 0, the roots are imaginary. 

This quantity, b 2 — 4 ac, is called the discriminant of the 
equation, and when placed equal to zero expresses the 
condition which must hold between the coefficients, if 
the two roots of the equation are equal. 



CHAPTER II 

THE POINT 

9. Cartesian coordinate systems. — The subject of Ana- 
lytic Geometry is, as its name implies, a treatment of 
Geometry by analytic or algebraic methods. It is then 
essential to have the means of translating geometric 
statements into algebraic and the reverse. Geometric 
theorems involve the ideas of magnitude, position, and 
direction. Algebraic methods of representing magnitude 
and direction have been considered in the previous chapter. 

The idea of position may be expressed algebraically in 
many ways. But at present Ave shall confine ourselves to 
two methods used in ordinary life. If we wish to locate 
a town, we usually speak of it as being a certain distance 
in a certain direction from some well-known location. In 
the plane we must have a fixed point A from which to 
measure the distance, and a fixed line AB from which to 
measure the direction of any 
point P. The point P is com- 
pletely determined when the 
angle BAP and the distance 
AP are given. This system 
of locating points in a plane is ' 

called the Polar System, and will be discussed fully later. 

Another method of fixing the position of a point on the 
earth's surface is to give its latitude and longitude, or its 




8 



ANALYTIC GEOMETRY 



[Cii. II, § 9 



II 



N>- 



'X— 



III 



->X 



IV 



distance north or south and its distance east or west from 
a given pair of perpendicular lines. 

Constructing a pair of perpendicular lines X' X and 
Y 1 Y in the plane, we may locate a point by saying that it 

is m units above or below 
X'X and n units to the 
right or left of Y'Y. If 
instead of using the words 
above or below, right or 
left, we understand that 
all distances measured up- . 
ward or to the right are 
positive, and those meas- 
ured downivard or to the 
left are negative, two num- 
bers with the proper signs 
attached will represent the distances of the point from the 
two lines, and these tivo numbers taken together will locate 
absolutely the position of any point in the plane. These 
numbers, representing the distances of the point from the 
two lines, with their proper signs attached, are called the 
coordinates of the point. The distance NP, measured from 
Y' Y, parallel to X'X, is called the abscissa, or jr-coordinate, 
and the distance MP, measured from X'X, parallel to Y' Y, 
is called the ordinate, or /-coordinate, of the point. The 
line X'X is called the axis of abscissas, or Jf-axis, and 
Y'Y the axis of ordinates, or K-axis. The two lines to- 
gether are called the axes of coordinates, or coordinate axes, 
and their intersection the origin of coordinates, or simply 
the origin. The abscissa of a point is denoted by the 
letter x, the ordinate by y, and the two coordinates are 



Y' 
Fig. 7. 



Ch. II, § 9] THE POINT 9 

written in a parenthesis (#, ?/), the abscissa being always 
written first. 

It will be seen at once that any point in the plane can 
be located by means of its coordinates, and that there will 
always be a point which will correspond to any pair of 
values we may choose, and that there will be only one 
such point. AVe have then a simple means of representing 
position in a plane by algebraic symbols. This system is 
called the rectangular, and is a particular case of Cartesian 
coordinates. In the general Cartesian system the axes 
are not necessarily perpendicular to each other. In case 
they are not perpendicular, the system is called oblique. 
All the definitions given above hold for the oblique 
system. 

In Fig. 8, 1VP is the abscissa of P and MP is its ordi- 
nate. While rectangular coordinates are more often used 
because their formulas are * Y 

simpler, yet it will occa- / 

sionally be desirable to use p JI / I 

J i <_ J N 

the more general svstem. / / 

But rectangular coordi- X~ /° * > x 

nates will always be un- / 

derstood unless another ni / IV 

system is distinctly speci- / 

fiecl. 

In locating or plotting a 
point whose coordinates are given, some convenient unit 
of measure must first be chosen. Then measure off the 
proper number of these units from the origin along each 
axis in the direction indicated by the sign of the coordinate. 
Through the points thus determined draw lines parallel 



Fig. 8. 



10 



ANALYTIC GEOMETRY 



[Cii. II, 



to the axes, and their intersection will locate the point 
whose coordinates were given. The following figures 
illustrate the method. Coordinate paper having two per- 
pendicular sets of parallel lines is very useful, and should 
be obtained by the student. 
*Y 



(■'*,?),--—'■ 



(S,-7)~ 



-JS,S) 



7,-5) 




Y 
Fig. 9. 



PROBLEMS 

1. Plot the following points : 

(0, 0), (0, - 3), (4, 0), (- 4, 0), (- 4, 5), (- 3 3 - 8). 

2. Construct the quadrilateral whose vertices are the points 
(7, 2), (0, - 9), (- 3, - 1), and (- 6, 4). 

3. TVhat relation exists between the coordinates of two 
points if the line joining them is bisected at the origin ? 

4. TVhat are the coordinates of the corners of a square 
whose side is s, if the origin is at the centre of the square and 
(a) the axes are parallel to the sides, (b) the axes coincide 
with the diagonals ? 

5. If one side of a parallelogram coincides with the X-axis 
and one vertex is at the origin, express in the simplest way 
the coordinates of the other vertices, («) in rectangular coor- 
dinates, (5) in oblique coordinates. 

6. What are the coordinates of the vertices of an equi- 
lateral triangle, if (a) one side coincides with the A^-axis and 



Ch. II, §§ 10, 11] 



THE POINT 



11 



the origin is at one vertex, (6) one side coincides with the 
X-axis and the origin is at the middle of this side, (c) if the 
origin is at the centre of the triangle and the X-axis passes 
through one vertex ? 

10. Notation. — It will often.be necessary to distinguish 
between points which are fixed and those which, although 
constrained to move in a certain path, yet can occupy 
any position along this path. Fixed points will always 
be distinguished by means of subscripts, being lettered 
P v P 2 , etc., and represented by the coordinates (x v y^), 
(# 2 , ?/ 2 ), etc., while variable points will generally be rep- 
resented by the simple variables (#, y). If variable 
points whose movements are governed by different laws 
are under consideration at the same time, they will be 
distinguished by using (x, ?/), (x 1 ',. y'), etc. 

11. Distance between two points in rectangular coordi- 
nates. — One of the first questions which naturally arises 
concerning points is that of finding the distance between 
them when their coordinates are given. 

Let P x and P 2 (in either figure) be two points whose 
coordinates are (x v y{) and (x v y 2 ~). Drop perpendicu- 



Mo 



Y 

Fig. 11. 



Mr 




12 ANALYTIC GEOMETRY [Ch. II, § 11 

lars on the X-axis and draw P 2 K to meet M 1 P 1 or that 
line produced. Then x 1 = 0M V x 2 = 0M 2 , y x = M 1 P V 
and y 2 = M 2 P 2 . It must be remembered that the coordi- 
nates of any point are measured from the coordinate axes 
and must be so read. 



In either figure P X P 2 = ^ I\K 2 + JS^ 5 ". 
But P 2 K= M 2 M X = 0M X - 0M 2 = x x - x 2 , 
and KP 1 = M 1 P l - M X K= M 1 P 1 - M 2 P 2 = y x - y 2 . 

Hence P x p 2 = y/<jxx-x*y + {vi-y»)\ [1] 

This being a length merely, it is immaterial whether 
it is read P X P 2 or P 2 P V 

It is necessary for the student to make himself familiar 
at once with demonstrations of this kind in which a single 
demonstration will apply to all possible cases. It might seem 
at first that in Fig. 12 the equation KP X = M 1 P 1 -M 1 K 
does not hold. But if M X K is replaced by its equal, 
— KM V the equation is at once seen to be true. 

Let the student draw various figures with the points 
in different quadrants, and assure himself that the same 
demonstration holds for all. Care must be taken to read 
the lines always in the proper direction. For simplicity 
the figures will usually be constructed in the first quad- 
rant, but the student should always satisfy himself that 
there is no restriction on their position, and that, if any 
other figure is constructed and lettered in a correspond- 
ing way, just the same demonstration will hold letter 
for letter. 



Ch. II, § 12] 



THE POINT 



13 



12. Distance between two points in oblique coordinates. — 
When the axes are oblique, draw M 1 P 1 and MJP^ the 
ordinates of P x and jP 2 , and the line P 2 K parallel to 
the X-axis. Since P 2 K and KP X are to be expressed 
in terms of the coordinates of P x and P 2 , their positive 




Fig. 13. 



Fig. 14. 



directions will be the same as the positive directions of 
the axes. The angle between them will always be &), 
and the generalized form of the law of the cosines 
(Art. 7) gives 



P X P 2 = Vp 2 JT 2 + KP X 2 + 2 P 2 K- KP X cos ft), 

where not only the magnitudes, but also the directions 
of the lines are considered. But 

P 2 K= M 2 M t = 0M 1 - 0M 2 = x x - x v 

and KP X = M l P 1 - M X K= M l P l - M 2 P 2 = y x - y v 

Substituting these values, we have 
P1P2 = V(oc! - x 2 )' 2 +(2/1 - y 2 ) 2 + 2(05i - x 2 XVi ~ 2/2) cos «, [2] 
as the distance between two points in oblique coordinates. 



14 ANALYTIC GEOMETRY [Ch. II, § 13 



PROBLEMS 

1. Find the distance between the two points whose rec- 
tangular coordinates are (—2, 6) and (1, 5). 

Solution. — In using formulas [1] and [2] we may choose either of 
the points for P x and the other for P 2 . Let (—2, 6) be the coordinates 
of Pi, and (1, 5) the coordinates of P 2 . 



Then P X P 2 = V( - 2 - l)' 2 + (6 - 5) 2 = v 10. 

2. Find the lengths of the sides of a triangle if the rec- 
tangular coordinates of its vertices are (— 3. 4), (— 6.— 1), and 
(4,-5). 

3. Find the lengths of the sides of the triangle, the coordi- 
nates of whose vertices, referred to axes making an angle of 
60° with each other, are (0, 0), (- 5, - 5), and (1, - 3). 

4. TYhat is the distance from the origin to the point (a, b) 
in rectangular coordinates ? In oblique coordinates, if the 
angle between the axes is 45° ? 

5. Show that the points (6, 4), (2, 8), (3, - 2), and (- 1, 2) 
are the vertices of a parallelogram. 

6. Show that the lengths of the diagonals of any rectangle 
are equal. 

Note. — Take the two adjacent sides as axes and call the opposite 
vertex (a, &). 

13. Points dividing a line in a given ratio. — The next 
question to be discussed is that of finding the coordinates 
of a point which will divide the line joining two given 
points in any given ratio. We must first define what we 
mean by " dividing the line joining two points in any 
given ratio " ; for it has a larger meaning here than we 
have been accustomed to give it. 

If C is any point on the line AB, it is said to divide the 



Ch. II, § 13] 



THE POINT 



15 



line AB into the two parts AC and CB (care being taken 
to read the two parts in just this way) whether the point 
C lies between A and B or 

beyond either. It will be £ - Q 

seen that it the point (J lies 

between vi and P, the ratio, -7=, of the parts into which 

it divides the line is positive ; while if it lies on AB pro- 

AC 

duced, the ratio is negative. If — — has a value between 

AC 

1 and — 1, C is nearer A, while if — — is greater than 1 

or less than — 1, C is nearer B. 

We shall now obtain the formulas for finding the coor- 
dinates of the point P which divides the line P X P 2 * n the 

4-1, 4. P^ »'l 

ratio m 1 : ra 2 , or so that — *— = — *- 




p 


F 


4-M 


T 


-^if if, x. 


\m 2 x 
^p 2 



Fig. 17. 

Draw the ordinates i^Pj, ilf 2 P 2 , and MP. Also draw 
the lines P 1 K 1 and PiT parallel to the X-axis. The 
triangles PP 1 K l and PKP 2 are evidently similar, and 

P X K X = K^P = P X P = rr^ 
PK KP, PP., m' 



16 ANALYTIC GEOMETRY [Ch. II, § 14 

But P X K X = OM - 0M X = x - x v 

PK = 0M 2 -OM ='x 2 -x, 
Kx P =M P -MK x =y -y v 
KP 2 = M 2 P 2 -M 2 K=y 2 -y. 

Substituting these values, we have 

3 = ^i, and y-yi M a. 



h - * »»2 Vi - y 

Solving, » = «*<■»+■—»» , ail d ,, *> + ■» . [8] 

If the point P bisects the line P X P V m l = m v and the 
formulas become 

x = *±±**, and y = V^. [4] 

Let the student go over the demonstration carefully, 
using the second figure, and assure himself that every 
step holds as well for that as for the first. Let him also 
construct other figures with the points in different posi- 
tions, but using the same letters for corresponding points. 

Since the demonstration depends only upon the simi- 
larity of triangles, it will hold also in oblique coordinates. 
The results are therefore general, and will apply to either 
system of Cartesian coordinates. 

14. Harmonic division. — If the line A C is divided by 
the points B and D, internally 
— . and externally, in the same 



numerical ratio, or so that 

— - = — — — , the line AC is said to be divided har- 
jd c u o 

monically. 



Cii. II, § 14] 



THE POINT 



17 



Let the student show that the line BD will then be 
divided harmonically by the points C and A. or so that 

BC = BA 
CD AD 

The four points A, B, C, and D are said to form a 
harmonic range. 

If parallel lines are drawn through the points A, B, (7, 
and D of a harmonic range, 
their intersections A\ B', C\ 
and D', with any transversal, 
will also be a harmonic range. 

For, from plane geometry, 



AB 
BC 



A'B' 

~~B'C r 

Hence 



and 



AD A'D f 



A 


B 


C 


A[^~~ — 


P/^ 


9^^ 



DC D ! C 

A'B 1 = 

B'C 



A'D' 
D'C 



Fig. 19. 



PROBLEMS 

1. Find the points of trisection of the line joining 
(- 3, - 4) and (5, 2). 

Solution. — If we wish to find P, the point of trisection nearest P 2 , 
mo = 1, mi = 2. 




_ m\X<2. + m 2 xi 
m\ + m 2 



_ 2.5 + l(-3) _7 
1 + 2 3' 



. fflij/a + ni2yi _ 2 . 2 + 1 (- 4) _ 



mi + m 2 



1 + 2 



and the point of trisection is (f, 0). 
But if we wish to find P', m\ = 1, m 2 = 2, 



1 


5 + 2(- 


3)_ 


1 


1 


1+2 
• 2 + 2(- 


■4)_ 


3 

2 




1 + 2 







and the other point of trisection is (— i, — 2). 



18 ANALYTIC GEOMETRY [Ch. II, § 14 

2. Find -the point which divides the line through ( — 3, —4) 
and (5, 2) in the ratio — §. 

3. Extend the line through (1, 5) and (—3, 4) beyond the 
latter point until it is three times its original length. Find 
the coordinates of its extremity. 

4. In the triangle whose vertices are (0, 0), (0, 6), (5, 8), 
find the point on each median which is two-thirds of the dis- 
tance from the vertex to the middle point of the opposite side, 
and show that these points coincide. 

5. Show that the medians of any triangle meet in a point, 
choosing the axes so that the vertices may be represented by 

(0, 0), (a, 0), and (b, c). 

6. In the right triangle whose vertices are (0, 0), (0, 6), and 
(8, 0), show that the distance from the vertex of the right 
angle to the middle point of the opposite side is equal to one- 
half of the hypotenuse. 

7. Prove that the theorem of problem 6 holds for any right 
triangle. 

Note. — Take the legs of the triangle as axes. 

8. In the triangle whose vertices are A (— 1, 2), B (4, 5), 
and C (3, — 4) , a line DE is drawn through the middle points 
of the sides AB and AC. Show that BC = 2 DE. 

9. Prove that the line joining the middle points of the sides 
of a triangle is equal to one-half of the third side, using the 
points (.Tj, 2/i), (x 2 y 2 ), and (x 3 , y s ) as the vertices of the triangle. 

10. If the coordinates of three of the vertices of a parallelo- 
gram are (0, 0), (8, 0), and (3, 5), find the coordinates of the 
fourth vertex, which lies in the first quadrant. 

11. Prove that the diagonals of any parallelogram bisect 
each other. 

12. In what ratio is the line joining the points (— 1, 6) and 
(7. - 2) divided by the point (2, 3) ? by the point (10, - 5)? 



Ch. II, § 14] THE POINT 19 

13. The line joining the points (0, 3) and (9, 0) is divided 
internally by the point (3, 2). Find the coordinates of the 
point which divides it externally in the same numerical ratio. 

14. Find the coordinates of the point P which forms, with 
the points A (4,1), B (2, — 2), and C (— 2, — 8), a harmonic 
range, if (a) P is between A and B ; (b) P is between B and 0. 



CHAPTER III 

LOCI 

15. Equation of a locus. — In the previous chapter we 
have considered fixed points only. If a point is made to 
move in the plane according to some definite law, a curve 
or locus is generated. (The term " curve " in Analytic 
Geometry is applied to any locus, including straight 
lines.) As, for example, a point which remains at a fixed 
distance from a given fixed point generates a locus called 
a circle ; a point which is always equally distant from 
two intersecting lines generates a locus which is the 
bisector of the angle between those lines ; a point which 
is always equally distant from the ends of a line generates 
the perpendicular bisector of that line, etc. 

If now we can translate the statement of the law govern- 
ing the movement of a •point into an algebraic relation or 
equation between the coordinates of the points which satisfy 
the laiv, we shall have an equation which may be used to 
represent the curve. For, if our translation is correct, 
every point whose coordinates satisfy .the equation will 
occupy a position on the path generated by the moving 
point, since the equation is only a restatement of the law 
itself in algebraic language. There will be, moreover, no 
position of the moving point whose coordinates do not 
satisfy the equation. We shall then have obtained an 
equation which is satisfied by the coordinates of every 

20 



Ch. Til, § 15] 



LOCI 



21 



M 



point on the locus, and by the coordinates of no point 
not on the locus. 

In the first example given above, if the fixed point is 
taken as the origin, and if the moving point P remains 
at a distance a from the fixed point, the relation between 
the coordinates x and y of every position of P is x 2 + y 2 = a 2 . 
For (Fig. 21) OM 2 + MP 2 = OP 2 . 

We see, moreover, that this 
equation cannot be satisfied by 
any point which is not at a dis- 
tance a from 0. We have then 

X— 

translated the given condition 
into algebraic language. The 
equation and the curve bear to 
each other the following recip- 
rocal relation : The coordinates 
of every point on the circle satisfy 

the equation, and conversely, every point ivhose coordinates 
satisfy the equation lies on the circle. When an equation 
and a curve are connected by this relation, the equation 
is spoken of as the equation of the curve, and the curve as 
the locus of the equation. 

Again, let a point move so as to remain equally distant 
from the two axes. What is the algebraic translation of 
this law, or, in other words, what is the algebraic equation 
which must be satisfied by the coordinates of every point 
governed by the law ? It is evidently x — y, and this is, 
therefore, the equation of the bisector of the angle be- 
tween OX and OY. 

What is the equation of the bisector of the angle 
between OF and OX'! 



Y 
Fig. 21. 



22 



ANALYTIC GEOMETRY 



[Ch. Ill, § 15 



If a point moves so as to be always three units above 
the Jf-axis, the ordinate of every point must be three, 
while no restriction is placed on the abscissa of the point. 
This law, translated into algebraic language, is, therefore, 
y = 3 ; for this equation makes just the same statement 
in regard to the position of every point which satisfies it. 
What is the equation of the locus of points two units 
to the left of the Y"-axis ? 

What are the equations of the axes ? 
The third illustration was the locus of a point which 
moves so as always to be equally distant from two fixed 
points. 

Place the axes with the origin at one of the points, and 
the X-axis coincident with the line joining the two points. 
Let the distance OA (Fig. 22) 
between the two points be rep- 
resented by a. Then the coor- 
dinates of the two points are 
(0, 0) and (a, 0). We can 
translate into an algebraic equa- 
tion the statement that a point 
P, coordinates (x, «/), shall be 
equally distant from the two 
fixed points and A, by ex- 
pressing the distances of P from each of the two points, 
and equating these two expressions. 




In Fig. 22, 
and 

Equating, 



OP = Vx 2 + ' 



AP = V(z - a) 2 4- f. 



~Vx 2 -f- y 2 = V(# — a) 2 + y l . 



By [i] 
By [i] 



Ch. Ill, § 15] LOCI 23 

Squaring and reducing, we have, as the equation of the 
desired locus, 



Here the final result does not express so clearly as in the 
previous cases that it is simply a translation of the state- 
ment of the law. But it has been obtained by simple 
algebraic reductions from this exact statement. The 
result is, as we should expect, the perpendicular bisector 
of the line joining the two fixed points. 

We have in these simple cases been able to translate 
the law governing the movement of a point in the plane 
into an algebraic equation. There are many loci for 
which this is possible. But the law may be stated in 
such a way as to require other than algebraic symbols to 
represent it. For example, the path of any fixed point 
on the circumference of a wheel rolling on a straight line 
in a plane is a perfectly definite curve. But the relation 
between the coordinates cannot be expressed in a single 
algebraic equation. It requires the introduction of trigo- 
nometric functions. 

If a point moves at random, no equation connecting the 
coordinates of its different positions can be found ; for an 
equation imposes a law upon the movement of the point. 

PROBLEMS 

1. Find the equation of the locus of points which are 
equally distant from the points (1, 3) and (—2, 5). 

2. Find the equation of the locus of points which are three 
times as far from the X-axis as from the F-axis. 

3. Find the equation of the locus of points Avhich are five 
units from the point (—3, 4). 



24 ANALYTIC GEOMETRY [Ch. Ill, § 16 

4. A point moves so as to be always five times as far from 
the F-axis as from the point (5, 0). Find the equation of its 
locus. 

5. A point moves so that the sum of the squares of its 
distances from the points (0, 0) and (o, — 5) is always equal to 
40. Find the equation of its locus. 

6. A point moves so as to be always three times as far 
from the point (1, — 2) as from the point (—3, 4). Find the 
equation of its locus. 

7. A point moves so that the sum of its distances from the 
two axes is always equal to 10. Find the equation of its locus. 

8. A point moves so that its distance from the X-axis is 
always one-half its distance from the origin. Find the equation 
of its locus. 

9. A point moves so that its distance from the point 
(—4, 1) is always equal to its distance from the origin. Find 
the equation of its locus. 

10. A point moves so that the square of its distance from 
the origin is always equal to the sum of its distances from the 
axes. Find the equation of its locus. 

16. Locus of an equation. — Looking at the question 
from the other side, let us consider what will be the 
geometric interpretation of any given equation in x and y. 
It is at once evident that only the coordinates of certain 
points in the plane will satisfy the equation; for, if we 
give any particular value to x, one or more values of y 
will be determined. The point, then, cannot occupy any 
position at random in the plane, yet it is not confined to 
a finite number of positions. For, since any value we 
please may be assigned to x, there will be an indefinite 
number of positions whose coordinates will satisfy the 



Ch. Ill, § 17] LOCI 25 

equation. Moreover, it appears that these points are not 
scattered indiscriminately over the plane, since random 
values of x and y will not satisfy the equation. Small 
changes in the value of x will in general produce small 
changes in the value of y. Points may therefore be 
found as close as we please to each other, and from this 
we may infer that they are situated on some curve. This 
curve which contains all the points which satisfy the equation 
and no others is called the locus of the equation. 

17. Plotting the locus of an equation. — How shall we 
determine the locus of any given equation ? Sometimes 
the locus is at once evident. For example, what is the 
geometric interpretation of the equation y — 3 ? The 
equation says nothing concerning the abscissas of points 
on the locus, but fixes the ordinate of every point. All 
points which satisfy it must therefore lie at a distance of 
three units above the X-axis. Hence the locus is a line 
parallel to the X-axis, and three units above it. 

Again, consider the equation x = y. It states in alge- 
braic language that a point moves so as to remain equally 
distant from the two axes. Its locus is therefore the line 
w r hich bisects the angle between the tw T o axes. 

Sometimes it is easy, as in these cases, to translate the 
algebraic equation into the law which governs the move- 
ment of the point, and hence determine the exact form 
and position of the locus. But this is often difficult, and 
we must have other means of determining the curve. We 
can always determine as many points as Ave please on the 
locus by giving to one of the coordinates a series of values 
and determining the corresponding values of the other. 



26 



ANALYTIC GEOMETRY 



[Ch. Ill, § 17 



Place these points in their proper positions in the plane, 
and when a sufficient number has been obtained, a smooth 
curve passed through them will show approximately the 
form of the curve. The points can be determined as near 
to each other as we please, and the approximation can be 
carried to any required degree of accuracy. This is called 
plotting the curve. 

We shall plot the locus of the equation 

2 x + y = 10. 

Give consecutive values to x, and find the corresponding 
values of y. 

If 




x = 0, 




y = 10. 


X = 1, 




y= 8, 


x=% 




y = 6, 


x = 3, 




y= 4, 


*=4, 




y= 2, 


x = 5, 




y= o, 


x = - 


1, 


y=i2, 


x = - 


2, 


^ = 14, 


x = - 


3, 


2/ = 16, 


etc 




etc. 



Plotting the points (0, 10), (1, 8), (2, 6), etc.. and 
passing a curve through the points, we see that they all 
appear to lie on a straight line. This method, however, 
does not assure us that the locus is a straight line. It 
only shows that, so far as our construction is accurate, it 
appears to be a straight line. 

AVe shall show later that every equation of the first 
degree represents a straight line. 



Ch. Ill, § 18] 



LOCI 



27 



Again, let us plot the locus of the equation 
x*-f= 25. 



Solving the equation for y, we have y = ± Vz 2 — 25, 
from which it appears that y is imaginary, so long as 
— 5 < x < -f- 5. There will therefore be no points on the 
locus for which x is numerically less than 5. 

If x = 5, y = ; x = — 5, y = ; 

a; = 6, */ = ±VTl; a = — 6, 2/ = ±VlT; 

x—1, y = ± V24 ; etc. 

Plotting the points (5, 0), (6, + VII), (6, - VTT), etc., 
and passing a smooth 
curve through them, we \ y 

have the curve in Fig. 
24. It can be seen from 
the equation that each 
branch goes off indefi- 
nitely, never again turn- 
ing toward either axis; 
for as x increases, y in- 
creases indefinitely. 

18. Symmetry. — A curve is said to be symmetrical 
with respect to one of two axes (rectangular or oblique) 
when that axis bisects every chord parallel to the other. 

A curve is said to be symmetrical with respect to 
a point when that point bisects every chord drawn 
through it. 

It is easily proved that if a curve is symmetrical with 
respect to two axes, it is symmetrical with respect to 




28 ANALYTIC GEOMETRY [Ch. Ill, § 18 

their point of intersection. Now, if, upon substituting 
any value for x in an equation, we find two values of #, 
equal numerically but with opposite signs, the curve is 
evidently symmetrical with respect to the X-axis. Or, 
if, for every value of y, we find two values of x, equal 
numerically but with opposite signs, the curve is evi- 
dently symmetrical with respect to the Y-axis. If both 
these occur, the curve must be symmetrical with respect 
to the origin. 

It appears that the first of these conditions can be 
satisfied when y occurs in the equation in even powers 
only, and the second when x occurs in even powers only. 
A curve- is therefore symmetrical with respect to the X-axis 
tvhen its equation does not contain odd powers of y ; it is 
symmetrical with respect to the Y-axis when its equation 
does not contain odd powers of x. 

It is symmetrical with respect to the origin if its equation 
contains no term of an odd degree in x and y. 

We can therefore tell at once whether a curve is sym- 
metrical with respect to either or both axes. This is 
useful in plotting ; for if a curve is sj^mmetrical with 
respect to the X-axis, it is only necessary to plot the part 
above that axis and form the same curve below ; if sym- 
metrical with respect to the Y-axis, to plot the part at 
the right of that axis and form the same curve at the 
left. 

The curve which we have just plotted, x 2 — y 2 = 25, 
is evidently symmetrical with respect to both axes. It 
would have been sufficient to have plotted that part which 
lies in the first quadrant and determined the rest of the 
curve from this. 



Ch. Ill, § 18] LOCI 29 

PROBLEMS 

1. Plot the loci of the following equations : 

(a) x 2 + y 2 = V. (/) 4ae?+-9y* = 0. 

(b) x* + y 2 = 0. (g) 4:X 2 -9y 2 = 0. 
( C ) a*- 2/2 = 0. (h) y 2 = ±x. 

(d) 4^ + 9?/* = 36. (i) x 2 = 4y. 

(e) 4:X 2 -9y 2 = 36. (J) y 2 = -4:X. 

2. Plot the locus of the equation 

£C 2 + 2y 2 -4« + 4y-12=0. 

The form of the equation shows at once that the curve is 
not symmetrical with respect to either axis. Solving for x in 
terms of y, we have 

tf _ 4 x = 12 - 4 y - 2 y 2 , 



or x = 2± Vl6-4?/-22/ 2 . 

From which it appears that the locus is symmetrical with 
respect to the line x = 2. 

There will be real values of x only for those values of y which 
make 16 — 4 y — 2 y 2 positive or zero. This expression vanishes 
when y = 2, or — 4, and can be factored into (4 -f- y) (4 — 2 y). 
It is evidently positive when — 4 < y < 2, and negative for all 
other values of y. 

Solving the given equation for y in terms of x, we have 



-.±v* 



_ ,-4a;- 
2/ = 

From which it appears that the locus is symmetrical with 
respect to the line y = — 1. 

The expression 14 + 4aj — x 2 vanishes when x = 2 ± 3V2, 
and can be factored into (2 + 3 V2 — x) (— 2 -f- 3 V2 + aj). It is 
evidently positive when 2 — 3 V2 < a; < 2 + 3 a/2, and negative 
for all other values x. There are then real points on the locus 
only when 2 - 3 V2 < x < 2 + 3 V% and - 4 < ?/ < 2. 



30 



ANALYTIC GEOMETRY 



[Ch. Ill, § 18 



The curve is therefore symmetrical with respect to the two 
lines x = 2 and y = — 1, and lies wholly within the four lines 
z=2-3V2, x = 2 + 3V2, y = -±, and y = 2. 

Giving y the values — 4, — 3, — 2, — 1, 0, 1, and 2, we have 
the following points on the locus : 

(2,-4), (2±VlO, -3), (6,-2), (-2,-2), 

(2 ± 3 V2, - 1), (6, 0), (- 2, 0), (2 ± VlO, 1), (2, 2). 

Plotting these points and drawing a smooth curve through 
them, we have a fairly clear notion of the form of the locus. 





























































































































































































































































































r - 




Fig. 25. 




Plot the loci of the following 


equations : 


(a) x 2 + y* + 2x-2y-10 = 0. 


(g) x- = if. 


(b) x 2 -\-y 2 -x-S = 0. 


(h) y = x\ 


(c) x* + y 2 — 4 # + 15 = 0. 


(?') y = sin x. 


(d) 4ar 2 + 9 2 / 2 -8 2 / + 6=:0. 


(j) y = tan x. 


(e) xy = 0. 


(k) y = sec x. 


(f)xy= 100. 


(7) y = cos -1 a- 



Cii. Ill, §§ 19-21] 



LOCI 



31 



19. Intercepts. — The distances from the origin to the 
points where a curve cuts the axes are called the intercepts 
of the curve. 

One of the coordinates of such a point will always be 
zero and the other will be the intercept. Hence the in- 
tercepts on the X-axis can be found by substituting 
y = in the equation and finding the corresponding values 
of x ; the intercepts on the 1^-axis, by substituting x = 
and finding the corresponding values of y. 

20. Intersection of two curves. — When two curves in- 
tersect, the coordinates of the point of intersection must 
satisfy both equations. In order to find the coordinates 
of such a point of intersection, it is only necessary to find 
the values of x and y which will satisfy both equations, or 
in other words, to solve the equations simultaneously. 



21. Locus of u -f kv =o and nv = o. — If all the terms 
of an equation are transposed to the first member, we may 
represent them by a single letter, as u or v, and speak of 
the equation as u = or 
v = 0. The letters u and 
v are simply used as ab- 
breviations for expres- 
sions in x and y of any 
degree. Then u = will 
represent some curve, and 
v = another curve. Let 
us consider what will 
be represented by the 
equation u + kv = 0, where Fig. 26. 




32 ANALYTIC GEOMETRY [Ch. Ill, § 21 

k is any constant quantity, positive or negative. Let 
(x v y x ) be any point of intersection of the two curves 
u = and v = 0. Its coordinates will satisfy both these 
equations, and hence will satisfy the equation u + kv = 0. 
The locus of u + kv = must therefore pass through all the 
points common to the two curves u = and v = 0. More- 
over, it will not pass through any other point of either 
curve. For the coordinates of any such point will cause 
one of the expressions u or v to vanish, but not the other, 
and therefore cannot satisfy the equation u -f kv = 0. 

Again, let us consider what will be represented by the 
equation uv = 0. It is evident that the coordinates of 
every point which cause either u or v to vanish will satisfy 
this equation, and that the coordinates of no other point 
can satisfy it. uv = must therefore represent the loci of 
the two equations u = and v = 0, taken together. P^or 
example, xy — represents both coordinate axes. 

PROBLEMS 

1. Find the intercepts of the curves whose equations are 
given on page 29. 

2. Find the points of intersection of the following curves : 

(a) x 1 + y- = 25 and x -f y = 4. 

(6) x 2 + y 2 = 25 and 3 x — 4 y = 25. 

(c) x 2 + y 2 = 25 and x + 2 y = 10. 

(c?) 3^ + 4^ = 24 and a 2 - ?/ 2 = 4. 

(e) ?/ 2 = 4 aj and # — i/ + 1 = 0. 

(/) tf 2 + 4 y 2 = 16 and 6y = x 2 . 

3. If the equations of the sides of a triangle are x + 7 y + 11 
= 0, 3 a + ?/ — 7 = 0, and x — 3^ + 1 = 0, find the length of 
each of the medians. 



Ch. Ill, §21] LOCI 33 

4. Which of the points (3, -1), (7, 2), (0, -2), and (8, 3) 
are on the locus of the equation 4 x — 7 y = 14. 

5. Find the length of the chord of intersection of the loci 
of x 2 + f = 13 and f = 3x + 3. 

6. For what values of b are the two intersections of the 
loci of y = 2 x + b and y' 2 = 4 x real and distinct ? imaginary ? 
coincident ? 

7. Write a single equation which will represent the two 
bisectors of the angles between the axes. 

8. Plot the two lines which are represented by each of the 
following equations : 

(a) x 2 + xy = 0. (c) 2x* 4- oxy - 3y 2 = 0. 

(b) x 2 -ox = -6. (d) 2y 2 -xy + ±x-9y = -±. 



CHAPTER IV 



THE STRAIGHT LINE 



22. We have seen that, if we know the law of the move- 
ment of a point, we can often determine the equation of 
its locus. We shall now proceed to the systematic study 
of a few such loci, beginning with the straight line. 

The two most common ways of determining the position 
of a line are to give either two points on it, or a single 
point and the direction of the line. If either of these sets 
of conditions is given, the line is fully determined, and we 
should be able to find the algebraic relation which must be 
satisfied by every point on it. 

23. Line through two points. — Let the line pass through 
the two points P v (x v y^), and P 2 , (a; 2 , y 2 ), and let P, 
(a;, ?/), be any point on the line. Draw the ordinates 
M 1 P V MP, and M 2 P 2 , and the line P^K parallel to OX. 
Then from the similarity of the two triangles P X LP and 




Ch. IV, § 23] THE STRAIGHT LINE 

Substituting these values, we have 



V - 2/i 2/2 



2/i 



OC — OCi Xz — OCi 



35 



[5] 



This is then the algebraic relation between the coordi- 
nates x and y of any point on the line and the constants 
x v t/ v x 2 , and ?/ 2 , and is therefore the equation of the line. 
It is called the two-point form of the equation of the 
straight line. 

Let the student show that this equation cannot be sat- 
isfied by the coordinates of any point not on the line. 

The student should here, and in all the following 
demonstrations, assure himself that the proof is perfectly 
general. Place the lines 
and points in different 
positions, being careful 
to give the same letter 
to corresponding points, 
and the demonstrations 
ought to hold, letter for 
letter. For example, try 
the following figure with 
the above demonstration, 
being careful to note that 

P x L = M x O + OM =OM -0M V 
LP =LM +MP =MP - ML, 
P l K=M 1 + OM cl =OM 2 -0M V 
KP 2 = KM 2 + M 2 P 2 = M 2 P 2 - M 2 K. 




36 ANALYTIC GEOMETKY [Ch. IV, §§ 24, 25 

24. Line determined by its intercepts. — If the two 

given points should be, in particular, the points where the 
line cuts the axes, or if, in other words, the intercepts a 
and b are given, the equation can be found easily by sub- 
stituting (a, 0) for (x v y^) and (0, b) for (x v y 2 ) in 
equation [5]. It becomes 

y - Q = b - 

x — a — a 

or reducing, — + ^ = 1. [6] 

This is called the intercept form of the equation of the 
straight line. 

Let the student derive equation [6] geometrically with- 
out using equation [5]. 

25. Oblique coordinates. — In obtaining these equations 
of the straight line we have made no use of the fact that 
the axes are perpendicular. The only idea used was the 
similarity of triangles, which will be true in oblique as 
well as rectangular coordinates. The results will hold 
therefore for both systems of Cartesian coordinates. 

PROBLEMS 

1. Find the equation of the straight line through the points 
(- 1, 5) and (6, 0). 

Solution. — In applying formula [5] either point may be chosen as 
Pi and the other as P 2 . Here let (6, 0) be P x and (- 1, 5) be P 2 . Sub- 
stituting in [5] j we have as the equation of the line 5 x + 7 y = 30. 

2. Find the equations of the lines through the following 
points and find the intercepts of these lines on the axes : 

(a) (- 5, 4) and (3, - 1). (c) (4, 2) and (4, - 2). 

(6) (0, 0) and (4, 3). (d) (3, 5) and (- 7, 5). 



Ch. IV, § 26] 



THE STRAIGHT LINE 



37 



3. Find the equation of the line whose intercepts are 3 
and — 1. 

4. Does the line joining the two points (6, 0) and (0, 4) pass 
through the point (3, 2) ? the point (—4, 5) ? 

5. "What condition must be satisfied if the point (x^ y{) lies 
on the line joining the points (a? 2 , y 2 ) and (x 3 , y 3 ) ? 

6. The line joining the points (6, 2) and (7, — 3) is divided 
in the ratio of 2 to 5. Find the equation of the line joining 
the point ( — 5, — 5) to the point of division. 

7. The coordinates of the vertices of a triangle are (2, 1), 
(3, —2), and (—4, —1). Find the equation of the medians, 
and show that the coordinates of the point of intersection of 
any two medians satisfy the equation of the third, and that 
the three medians therefore meet in a point. 

8. What are the equations of the diagonals of the rectangle 
whose vertices are (0, 0), (a, 0), (0, b), and (a, b) ? Find the 
point of intersection, and show that they bisect each other. 

9. What system of lines is represented by the equation 

-4-^=1, if we keep a constant and allow b to vary ? if we 

keep b constant and allow 
a to vary ? 

26. Line determined by 
a point and its direction. 

— If the second condi- 
tion mentioned in Art. 
22 be given, — a point on 
the line and the direc- 
tion of the line, — we can 
obtain its equation as 
follows : 

Let (x v y{) be the given point, and let the direction of 
the line be determined by the angle y which it makes with 




Fig. 29. 



38 ANALYTIC GEOMETRY [Cn. IV, § 27 

the positive direction of the JT-axis measured in the posi- 
tive direction of rotation. In the triangle KP X P, 

|J = taniOV>. 



But for all 


positions of P, 




KP = y-y v 




P 1 K= x - x v 


and 


tan KP X P = tan 7. 


Hence 


"i = tan 7 , 



x — x 1 
or y — y 1 — tan 7 • (x — x^). 

Tan 7 is called the slope of the line, and may be repre- 
sented by I. The equation 

y -yi = l(x-aci) [7] 

is called the slope-point form of the equation of a line. 
By comparing equations [0] and [7] we see that 

x i x \ 

27. Line determined by its slope and its intercept on the 

X-axis. — If the point through which the line is to 

pass lies on the Y-axis, its coordinates being (0, 5), [7] 

reduces to 

y-b = lx, 

or y = lx + b. [8] 

This is called the slope form of the equation of a line. 
Let the student derive equation [8] geometrically 
without using equation [7]. 



Ch. IV, § 28] 



THE STRAIGHT LINE 



39 



28. Oblique coordinates. — In deriving equations [7] 
and [8] we have made use of the fact that the axes are 
rectangular. A separate demonstration is therefore 
necessary in oblique coordinates. 

Using the same construction and notation as in Art. 26, 
it is again true that P^K = x — x x and KP = y — y v But 
the triangle KP X P is 
not right-angled, and in 
order to find the ratio 
between its sides we 
must make use of the 
law of the sines. Let 
the positive direction of 
P X P be taken along the 
terminal line of the an- 
gle 7. Then the angle 
formed by the positive 

direction of P X P with the positive direction of P X K is 
always 7, and the angle formed by the positive direction 
of KP with the positive direction of P X P is (&)— 7). 




Fig. 30. 



Hence, lg- = !LU!i = si " 7 , (See Art. 7) 

P X K x — x t sin (<w — 7) 



V - 2/1 = . S / IU y , (x - xi). 
sin (« - y) 



[9] 



If the coordinates of the given point are (0, 5), [9] 
reduces to 

y= . sin ? oc+b. [10] 

9 sin ( w -y) L J 

When a) = 90°, these two forms will be seen to reduce 
to the equations [7] and [8]. 



40 ANALYTIC GEOMETRY [Ch. IV, § 29 

PROBLEMS 

1. What is the equation of the line which passes through 
the point (—6, 6) and makes an angle of 60° with the X-axis ? 

2r Find the equation of a straight line if 

(a) 6 = 6 and y = 30°, 

(b) b = — 5 and y = tan" 1 f , 

(c) 6 = 3, y = 30°, and a> = 60°. 

3. Find the equation of the straight line through the inter- 
section of the lines 2 x — 3 y = 4: and 3 x — y = 5, and making 
an angle of 120° with the X-axis. 

4. What is the slope of the line whose intercept on the 
Y-axis is 5 and which passes through the point (3, — 1) ? 

5. What system of lines is represented by the equation 
y —lx + b if we keep I constant and allow b to vary ? if we 
keep b constant and allow I to vary ? 

29. General equation of the first degree. — We have 
found that the equation of every straight line given by 
any of the preceding conditions is of the first degree 
in both rectangular and oblique coordinates. It now 
remains to consider whether an equation of the first 
degree can represent any other locus. 

Every such equation is included in the general form 

Ax+By + C=0, 

where A, B, and C can have any values, positive, negative, 
or zero. 

If B^0, we can divide the equation by it, and trans- 
posing, we have 

A C 
y = — — x — — * 
J B B 



Ch. IV, § 29] 



THE STRAIGHT LINE 



41 



where — — and 
B 



can have any value. But the slope 

form of the equation of a line has been shown to be 

y = lx + b. (Arts. 27, 28) 

Since I = tan y or in oblique coordinates — ^- 



sin ( co — 7)J 

and b is the intercept on the I^axis, they can have any 
real value whatever. 

We have then reduced the general equation 
Ax + By + C = 
to the slope form of the equation of a line, and it must 
represent that line for which 



I = — — and b = 
B 



C 
B 



UB-. 



■■ 0, the general equation reduces at once to 


X =~A 



which we know to be the equation of a line parallel to the 
y-axis. 

We have then shown 
that the general equation 
of the first degree always 
represents a straight line. 

Another method of 
showing that the locus 
of any equation of the 
first degree is a straight 
line is as follows : FlG * 31, 

Let (x v 2/j), (x v ^ 2 ), and (# 3 , y 3 ) be the coordinates of 
any three points on the locus of the equation 
Ax + By + C=Q. 




42 ANALYTIC GEOMETRY [Ch. IV, § 30 

These coordinates must satisfy the equation. Substi- 
tuting, we have 

(1) Ax l + By 1 + C=^ 

(2) Ax 2 + By 2 + C=Q, 

(3) ^3+% 3 + tf=0. 
Subtracting (2) from (1), we have 

- A (x 2 - x x ) = B(y 2 - y,), 

or % ~ x i = £ 

#1 ~ #2 A 

Subtracting (3) from (2), we have 

- A (x z - x 2 } = B (y 3 - y 2 ), 



Hence, 



The triangles KP^P 2 and LP 2 P S are therefore similar, 
and P X P 2 P Z is a straight line. 

30. Two equations representing the same line cannot 
differ except by a constant factor. — Let A x x + B x y + 6^ = 

and A 2 x + B 2 y + C 2 = represent the same line. Their 
intercepts on the axes must be the same. Hence, 

_ <?i = _ ^2, or ^ = —i, 

-/jL-i -*-*-9 ^'9 -"-9 

and _JZl = _^ or ^ = =?1. 

-^1 -^2 2 2 





*^3 *^2 - 

V2-VZ 


B 
= A 




#2 " 
2/1- 


- x 1 _ x% — x 2 

-V2 V2- y* 


KP 9 

or « 


XP 3 
XP 2 



Ch. IV, § 30] THE STRAIGHT LINE 43 

ABC 

Hence, — ^ = — i = — 1 , and the equations differ only by a 
A 2 B 2 6 2 

constant factor. 

The converse is easily seen to be true : if two equations 
of the first degree differ only by a constant factor, they rep- 
resent the same straight line. 

PROBLEMS 

1. Find the values of a, b, and I for the line whose equation 
is2z + 32/-12 = 0. 

Solution. — The intercepts are found, as explained in Art. 19, to be 
a = 6 and b = 4. The slope I may be found by changing the equation 
into the slope form, as explained in Art. 29. Transposing and dividing 
by 3, we have y = — § x + 4. Hence I = — f . 

2. Find the values of a, b, and I for the lines represented 
by the following equations, and construct the lines first by the 
aid of the intercepts a and b, then by the aid of the slope I, 
and the intercept b : 

(a) x-4:y -10 = 0, (c) 4x + y = 0, 

(b)3x-5y+ 7 = 0, (d) 2x + 8 = 0. 

3. Determine the values of A, B, and C, if the line 
Ax + By+C = passes through the points (3, 0) and (2, — 1). 

Solution. — Since the line is to pass through these points, their coordi- 
nates must satisfy its equation. By substitution, we obtain 

3.4+ C = 0, 

and .2 A - B + C = 0, 

two equations in A, B, and'O, from which the values of two of them may 
be obtained in terms of the third. Solving, we have C = — 3 A and 
B = — A. The equation of the line is therefore 

Ax-Ay-3A.= 0, 

or x — y — 3 = 0. 



44 



ANALYTIC GEOMETRY 



[Ch. IV, § 31 



4. Find by the same method the equations of the lines 
through the points 

(a) (3, 1) and (-5,0), 

(b) (0, - 2) and (3, 4), 

(c) (0, 0) and (5, - 3). 

5. Show that if two lines are parallel, their slopes must be 
equal ; if perpendicular, the slope of one must be the negative 
reciprocal of the slope of the other, [tan y = — cot(y + 90°).] 

6. Select pairs of the following equations which represent 
(a) parallel lines, (b) perpendicular lines : 

2x-Sy = 6, a = -f2/ + 6, 



4 x — 6 y = 7, 
12a + 8y = ll, 



V- 



-4y = 10. 



31. The angle which one line makes with another. — We 
have defined the angle between two directed lines as the 
angle between their positive directions. But when the 
lines are given by their equations and no convention is 

used to fix their posi- 
tive directions, it is con- 
venient to define the an- 
gle which one line makes 
with another as the angle 
formed by going from the 
second line to the first in 
the positive direction of 
rotation. This definition 
always gives a definite 
angle. - 

In Fig. 32, the angle which AB makes with MN is 
Z NOB, or its equal Z MCA ; the angle which MN makes 
with AB is Z BCM or Z ACN Let it be required to 





Y 


B 
XV 


N 






x^ -- -X 


.-~~.J« 




/^ 





\ w 


X 








w 









Y' 
Fig. 32. 



Ch. IV, § 31] THE STRAIGHT LINE 45 

find the angle which AB makes with MN. Let the equa- 
tions of MN and AB be given in the form 

y = l ±x + h v 

and y =J 2 x + b 2 . 

From the figure 6 = y 2 — y v 

and tan 6 = tan ( 7 , - 7l ) = fan 7 , - tan 7 , , 

1 + tan 7 2 tan y x 

Hence tan 9 = ~~- [11, a] 

If the equations are given in the form 
A l x + B l y + ^ = 0, 
and A 2 x + i? 2t y + C 2 = 0, 

it was shown in Art. 29 that 



h 


= -Aand 


h = ~ 


A. 
B. 




4„ n q _AiB 2 


- A 2 B X 






A.\JL% 


+ BiB, 






PROBLEMS 





Hence *** = ?,7.Z%Z: PM] 



1. Find the angle which the line ox — oy = 10 makes with 
the line a; + 2 ?/ = 7. 

Solution. — The angle is to be measured from the last line, and that 
line therefore takes the place of MN in Fig. 32. Hence l x = — |, and 
? 2 = f. Substituting these values in [11, a], 



; + ^ 



= 13, 



1 - 
or 6 = tan -1 13. 

If the question is reversed, and we wish to find the angle which the 

line x + 2 y = 7 makes with the line 5 £ — 3 y = 10, we must take h = f 

and ? 2 = — i- Then 

tang= ~~ *~ » = - 13, 
1-4 

or = tan- 1 (— 13). 



46 ANALYTIC GEOMETRY [Ch. IV, § 32 

2. Find the angle which the line 3x + 5y — 1 = makes 
with the line 11 x — 2 # + 3 = 0. 

3. Find the interior angles of the quadrilateral whose ver- 
tices are (3, 3), (5, - 3), (4, - 5), and (- 3, 0). 

4. The equations of the sides of a triangle are x-\-8 y+11 = 0, 
2 x — 3 y + 1 = 0, and Ax -f oy + 6 = 0. Find one exterior 
angle of the triangle and the two opposite interior angles. 

32. Perpendicular and parallel lines. — If two lines are 
parallel, tan 6 = 0, and therefore l x — Z 2 = 0, or 

This appears also from the figure, since if two lines are 
parallel, they must make the same angle with the X-axis. 

If two lines are perpendicular, tan 6 = oo, and therefore 
1 + y 9 = 0, or 

*--£ or 4-4- [13] 

This appears also from the figure, since if the lines are 
perpendicular, 

7i = 7 2 - f' and tan 7i = - cot 7 2 < or h=-T' 

— In 



l 2 

If now we wish to obtain the equation of a line parallel 
to a given line Ax + By + C= 0, the only condition which 
niust be satisfied is that it shall have the same slope. 
This can be accomplished by writing the equation 
Ax + By = k, where k is arbitrary. This will include all 
lines parallel to the given line, for by varying k it can 
be made to represent any one of the indefinite number 
of such lines. The value of h in any particular problem 
must be determined by some other condition. For 
example, if it is to pass through a given point, k can be 



Ch. IV, § 32] THE STRAIGHT LINE 47 

determined from the fact that the coordinates of this 
point must satisfy the equation. 

Again, if we wish to obtain the equation of a line per- 
pendicular to the line Ax 4- By + (7=0, we must write 

A B 

an equation such that — 1 = -. Such an equation is 

B 1 A 

Bx — Ay = Jc. This again contains all the lines perpen- 
dicular to the given line. If the line is to pass through 
a given point, k can be determined as before by the fact 
that the coordinates of this point must satisfy the 
equation. 

PROBLEMS 

1. Write the equations of the lines through (3, 4) which 
are respectively parallel and perpendicular to the line 
3aj-5y = 10. 

Solution. — The equation of the line which is parallel will be of the 
form Sx — 5 y = k. Substituting (3, 4), we have k = — 11, and the equa- 
tion of the parallel line is3x — 5 y = — 11. 

The equation of the perpendicular line will be of the form 5x+3y=k. 
Substituting (3, 4), k = 27, and the equation of the perpendicular line is 
5 x + 3 y = 27. 

2. Find the equation of the line through (5, 8) perpendicular 
to 3 x -f- 7 y = 21. 

3. In the triangle whose vertices are (0, 0), (6, 0), and (4, 8), 
find (a) the equations of its sides ; (b) the equations of perpen- 
diculars from the vertices upon the opposite sides ; (c) the 
equations of the perpendicular bisectors of the sides ; (d) the 
equations of the medians. 

4. Show that in the above problem the perpendiculars from 
the vertices, the perpendicular bisectors, and the medians each 
meet in a point. 

5. Show that the points obtained in problem 4 lie on a line, 
and obtain the ratio of the distances between them. 

6. Show that in any triangle the medians meet in a point. 



48 



ANALYTIC GEOMETRY 



[Ch. IV, § 33 



Note. — Choose the axes of coordinates so that the origin is at one 
vertex and the X-axis is coincident with one side of the triangle. The 
coordinates of the vertices of the triangle may then be taken as (0, 0), 
(a, 0), and (ft, c). 

7. Show that in any triangle the perpendiculars from the 
vertices on the opposite sides meet in a point. 

8. Show that in any triangle the perpendicular bisectors of 
the sides meet in a point. 

9. Show that the three points obtained in problems 6, 7, 
and 8 lie on a line, and find the ratio of their distances from 
each other. 

10. Show that the line joining the middle points of the 
sides of a triangle is parallel to the third side and equal to one 
half of it. 

11. Show that the diagonals of a square or rhombus are 
perpendicular to each other. 



33. Line making a given angle with a given line. — In 

plane geometry it is usual to speak of two lines through 

any given point and making a given angle with a given 

line. But if Ave con- 
sider the direction of 
the angle, there can be 
only one such line. For 
if MN is the given line, 
Pj the given point, and 
(f> the given angle, there 
can be only a single line 
which passes through P 1 
and makes the angle <f> 

with MN, where <f> is measured in the positive direction of 

rotation. 




Ch. IV, § 34] THE STRAIGHT LINE 49 

Let MS be this line. Let the inclination of MN be y v 
and of MS be y 2 . Then from [11, a J, 

Solving for ? 2 , Ave have 

j _ l r + tan (j> . 



2 " 1-^tanc/) 
The equation of MS will therefore be 

If MS is parallel to 3/iV, tan (f> = 0, and the equation 
becomes 

If MS is perpendicular to J/X. tan cf> = oo, and the 
equation becomes 

These formulas might be used to write the equations of 
parallels and perpendiculars in place of the methods given 
in the previous section. 

PROBLEMS 

1. Find the equation of the line through the origin which 
makes an angle of 60° with the line x — 3 y = 10. 

2. Find the equation of the line through (1, 4) which makes 
an angle of 135° with the line joining (1, 4) with the intersec- 
tion of 5 x — 2 y = 17 and 3 x -f 4 y = 5. 

34. Normal form of the equation of a straight line. — 

If we have given the length of the perpendicular or nor- 
mal from the origin on a line, together with the angle 



50 



ANALYTIC GEOMETRY 



[Ch. IV, § 34 




which this normal makes with the positive direction of the 
X-axis, the line is completely determined. The perpen- 
dicular distance is represented by p, and the angle by a. 

Through draw a line 
making an angle a with 
OX. If any distance OH 
is laid off on this line 
either in the positive di- 
rection (along the termi- 
nal line of the angle), or 
in the negative direction, 
and through H a line 
AB, perpendicular to OH, 
is drawn, that line is com- 
pletely determined. It 
is convenient to restrict a to positive values from 0° to 
360°. In case we wish to speak of a complete set of parallel 
lines without changing «, it will be necessary to allow p to 
be either positive or negative, but every line in the plane 
can be determined by positive values of both a and p, and 
this will always be understood unless otherwise stated. 
We have seen that the equation of the line AB in terms 

of its intercepts is - -f- j- = 1. But for all positions of the 

line 



and 



Fig. 34. 



p 

- = cos «, 
a 


P 

or a = 

cos a 


p 

f = sm «, 




or h = J^. 
sin a 



Substituting these values of a and 6, the equation of AB 

becomes 

ac cos a + y sin a = p. [15] 



Ch. IV, § 35] THE STRAIGHT LINE 51 

This is called the normal form of the equation of a 
straight line. 

Let the student show that the equation of a straight 
line in oblique coordinates in terms of « and p is 

x cos a -j- y cos (<w — «) = p. 

p p 

Note. — The equations - = cos a and - = sin a are true for all cases, 

a b 

since if p is positive, a and cos a have the same sign, and also b and sin a. 
While if p is negative, they have the opposite signs. 



PROBLEMS 
1. What is the equation of the straight line in which 

(a) a = 60°, and p = 5 ? (d) a = 225°, and p = ? 

(b) a = 120°, and p = 5 ? (e) a = 45°, oi = 60°, and p = 1 ? 

(c) a = 380°, and p= - 5 ? (/) a=- 60°, o> = 135°, and p=6 ? 

35. Reduction of the general equation to the normal 
form. — Since the general equation of the first degree 
Ax + By 4- O = always represents a straight line, it 
ought to be possible to reduce it to any one of the stand- 
ard forms. We have already shown how to reduce it to 
the slope form, and that 

= __ am = __. 

The following method enables us to reduce it to the 
normal form. If the two equations Ax + By + C=0 
and x cos a + y sin a — p = are to represent the same 
line, it was shown in Art. 30 that they can differ only by 
a constant factor. Let k be the quantity by which it is 



52 ANALYTIC GEOMETRY [Ch. IV, § 35 

necessary to multiply Ax -f By -f C = to make it iden- 
tical with x cos a + y sin a — p = 0. 

Then JcA = cos «, &i? = sin «, and JcC = — p. 

Squaring the first two and adding, we have 

k 2 A 2 + k 2 B 2 = cos 2 a + sin 2 a = 1. 

Hence & = ± — , and the equation 

V^L 2 + B 2 

[16] 



is then identical with x cos « 4- ?/ sin « — p = 0. 

If a andp are so chosen that p shall always be positive, 
then in any numerical case that sign must be given to the 

C 
radical which will make - a negative number 

±V^ 2 + ^ 2 
to correspond to — p. Hence the sign of the radical must 
be chosen opposite to the sign of C. This will always be 
understood unless the contrary is stated. 

PROBLEMS 

1. Reduce the equation 3 x 4- 4 y = 10 to the normal form. 



Solution. — Here ± VA 2 + B 1 = ± 5, and since C is negative, we must 
divide by + 5, and the equation becomes § x + § y = 2. 

Hence cos a — §, sin a = f, and p = 2. The line can be easily plotted. 
What would have been the values of a and p, if — 5 had been chosen ? 

2. Reduce the following equations to the normal form and 
plot the lines which they represent : 

(a) 4 a -3y = 25, (d) x + 4 = 0, 

(b) x + 2y = -$, (e) 5y -3 = 0, 

(c) 2x - y = 0, (/) x- 3y + 4 = 0. 



Ch. IV, § 36] 



THE STRAIGHT LINE 



5:3 



3. What system of lines is represented by the equation 
x cos a -\-y sin a — p = 0, if we keep a constant and allow p to 
vary ? If we keep p constant and allow a to vary ? 

36. Distance of a point from a line. — Let it be required 
to find the distance of the point P x from the line AB when 
the equation of AB is given in the form 

x cos a + y sin a — p = 0. 





Draw MN through P 1 parallel to AB and continue the 

perpendicular Off to meet it at K. The equation of MN 

will be 

x cos u + y sin a — p x = 0, 

where p x may be either positive or negative. For, as the 
value of a is fixed and as MN can be any line parallel to 
AB, it may be on the opposite side of the origin from AB, 
and in this case p x will be negative. (See Art. 34.) 

Since P x lies on MN, its coordinates (x v y-^) must satisfy 
the equation of MN. 



54 ANALYTIC GEOMETRY [Ch. IV. § 36 

Hence x x cos a + y x sin « = p v 

Now wherever P 1 may lie, RP X = HK= OK— OIT—p 1 —p. 

Hence B Pi = x x cos a + y x sin a -p. [IT, a] 

If the equation is given in the form Ax + By +(7=0, 
it is necessary first to reduce it to the normal form and 
then substitute x 1 for x and y 1 for y. 

Hence BP L = ^ ■ [17, 5 ] 

7%e radical must be given the sign opposite to that of C. 

It appears from the way BP X has been chosen that the 
result will be positive when the point and the origin are on 
opposite sides of the line ; negative, when they are on the 
same side of the line. 

PROBLEMS 

1. Find the distance of the point (3, 5) from the line 

2x-3y + 6=0. 

2. Find the distance of the origin from the line 

3 x + 4 y — 5 = 0. 

3. Find the area of the triangle whose vertices are (0, 3), 
(4, 0), and (5, 5) by calculating the length of one side and the 
distance of the opposite vertex from that side. 

4. Given the line 3 x — 4 y = 10 and the point (—3,5). 
Find the equation of the line through the point perpendicular 
to the given line ; find the point of intersection of this perpen- 
dicular with the given line ; find the distance of the given point 
from this point of intersection. 

5. Use the method indicated above to find the distance from 
the point (.r l5 y x ) to the line Ax + By -f- C = 0. 



Ch. IV, § 37] THE STRAIGHT LINE 55 

6. Find the distance between the two parallel lines 

7 x — 8 y = 15, and 7 x — 8 y = 40. 
Which line is nearer the origin? 

7. Show that the point (3, 1) is on the same side of the line 
x -f 4 y = 8 as the origin. 

37. Oblique coordinates. — It will be noticed that sec- 
tions 31-36 have reference to rectangular coordinates 
only. The corresponding formulas in oblique coordinates 
are rather complicated and seldom used. ^\ T e shall simply 
state what they are without obtaining them. 

To reduce Ax + By + C = to the normal form, 
x cos a -f y cos (to — a) = p, multiply the equation by 



VA? + B 2 - 2 AB cos o> 



The angle between two lines whose equations in oblique 
coordinates are A x x + B x y -f C x = and A 2 x + B. 2 y -f C 2 = 
is 

tan q 04-1^2 - ^jgQsina) 

The condition for parallelism is the same as in rectangu- 
lar coordinates, — 1 = — a . But the condition for perpen- 

A ^2 

dicularity is 

A^ + ^^ - (A^g -f- A 2 B^ cos to = 0. 

If, then, only parallel lines enter into a problem, oblique 
coordinates may be used with advantage ; but if it is 
necessary to use perpendicular lines, oblique coordinates 
should be avoided- 



56 



ANALYTIC GEOMETRY 



[Cii. IV, § 38 



The equation of a line perpendicular to Ax + By+ 0=0 

(B — A cos ft>) x — (A — B cos (t>)y = k. 
The distance of a point from a line is 

x x cos a -h y l cos (ft) — «) — jt?, 
(Ax, + By, + C) sin ft) 



VA 2 + # 2 - 2^L£ cosft) 



• Let the 



38. Bisector of the angle between two lines. 
equations of the two lines AB and MNbe 

(1) A x x + B,y + C\ = 0, 
and 

(2) A 2 x + B 2 y+C 2 =0, 

and let (V, y') be any 
point on the bisector of 
the angle between them. 
Since every point in the 
bisector of an angle is 
equally distant from the 
sides, HP' and KP' are numerically equal. But 

Kp , = a x x' + By + c\ and Hp , = A 2 x> + By + c\ 

±^A* + B* ' ±VA 2 2 + B 2 2 

Hence the relation which must exist between x' and y' 
in order that P' may be a point on the bisector is 




Aix' + Biy' + Ci _ Aooc' + B 2 y' + C 2 < 



[18] 



If the signs of the denominators have been chosen in 
accordance with the rule given in Art. 36, the positive 
si<ni in the second member indicates that P' and the 



Ch. IV, § 39] THE STRAIGHT LINE 57 

origin are either on the same side or on opposite sides 
of each of the lines, and that therefore the equation repre- 
sents the bisector of the angle in which the origin lies ; 
while if the minus sign is chosen, it represents the bisector 
of the angle in which the origin does not lie. 

If either Cj or C 2 is zero, one or both of the lines pass 
through the origin, and this test cannot be used. 

PROBLEMS 

1. Find the equations of the bisectors of the angles between 
the two lines 3 x — 4 y = 10 and 4 a,* + 3 ?/ = 7. Show that the 
two bisectors are perpendicular. 

2. Show that the bisectors of any pair of supplementary 
adjacent angles are perpendicular to each other, using the two 
lines in Art. 38. 

3. The equations of the sides of a triangle are 3x = 4y, 
4x = — 3y, and y = 6. Show that the bisectors of the interior 
angles meet in a point. Show also that the bisector of the 
interior angle at one vertex and the two bisectors of the 
exterior angles at the other vertex meet in a point. 

39. Lines through the intersection of two given lines. — 
If the equations of two given lines are 

(1) ^ r i- + £^4- 6\ = 0, 

and (2) A 2 x + B 2 y + C 2 = 0, 

and we form the equation 

(3) A t x + B x y + C\ + k(A 2 x + B 2 y + C 2 ) = 0, 

where k can have any value, it will represent for every 
value of Jc some line through the intersection of the first 
two. For the coordinates of the point of intersection 
of the loci of (1) and (2), which must satisfy both of 



58 ANALYTIC GEOMETRY [Ch. IV. § 39 

these equations, must satisfy (3) also. Moreover, it 
represents any line through their intersection, for k 
can always be chosen so as to make the locus of (3) 
pass through any given point. It is only necessary to 
substitute the coordinates of the point in the equation 
and determine k so that the equation is satisfied. In 
this way the equation of the line through the intersection 
of two lines and any other point may be obtained without 
actually finding the coordinates of the point of intersection. 
If any other condition sufficient to determine the line 
is given (for example, its slope), k can always be deter- 
mined so that the line will satisfy the condition. 

PROBLEMS 

1 . What is the equation of the line through the intersection 
of 2 x + 3 y - 4 = and x + 2y-5 = 0, and the point (2, 3) ? 

Solution. — The equation of any line through the intersection of the 
given lines is 

2 x + 3 y — 4 + k{x + 2 y — 5) = 0. 

Since the line is to pass through the point (2, 3), these coordinates 
must satisfy the equation. 

Hence k = — 3. 

Substituting this value, we have 

x + 3 y - 11 = 0, 
as the equation desired. 

2. "What is the equation of the line passing through the 
origin and the intersection of the lines x + 3 y — 8 = and 
4.r-5?/ = 10? 

3. In the triangle whose sides are 

5x-6y = 16, Ax + oy = 20, and x + 2y = Q, 

find the lines through the vertices and parallel to the opposite 
sides without finding the coordinates of the vertices. 



Ch. IV, § 40] 



THE STRAIGHT LINE 



59 



4. Find the equation of a line through the intersection of 
the lines 2x — 3 ?/ + 1 = and x -f- 5 y + 6 = 0, which is per- 
pendicular to the first of these lines. 

5. Find the equation of the line through the intersection 
of the lines y = 7 x — 4 and y = — 2 x -\- o, which makes an 
angle of 60° with the X-axis. 

6. Find the equation of the line through the intersection 
of the lines 5 y — 2 x — 10 = and y + 4 x — 3 = 0, and also 
through the intersection of the lines 10 y + x + 21 = and 
3y-5x + l = 0. 

40. Area of a triangle. — If the coordinates of the 
vertices of a triangle arc given, the area of the triangle 
may be found in the fol- 
lowing manner : 

The area is equal to 
the numerical value of 



JiTPgXP^. 



i\P 2 




Fig. 38. 



V(^ - x 2 y + (^ - y 2 y. 

HP Z is the distance of 
P 3 from the line P X P 2 . 
The equation of P X P 2 is 

(y% - y\) x ~ O2 - x i)y - x \y<i + *#i = °- 

, 0/2 ~ ffl>8 ~ 0*2 ~ A 'l)i/3 ~ ^2 + 3^ 
A2/2-yi) 2 +(^2-^l) 2 

and the area = i\_<iy 2 -y 1 )x s -(x 2 -x 1 )t/ s -x 1 y 2 +x 2 y 1 ], 

= il(^ 1 -x-2)tjs+ (xo-x s )yi + (iX3-Xi)ij2^. [19] 

The form of the result is easily remembered since the 
subscripts follow the cyclic order. The sign of the result 



60 ANALYTIC GEOMETRY [Ch. IV, § 40 

may be disregarded, since it is only the numerical value 
of the area we wish. 

The area of a polygon may be found by dividing 
it into triangles and finding the area of each triangle 
separately. 

PROBLEMS 

1. Find the area of the triangle whose vertices are (1, — 3), 
(- 4, 3), and (5, 5). 

2. Show that the area of any quadrilateral is 

3. What is the area of the quadrilateral the equations of 
whose sides are x = Q, x-\-y=0, x+2y=5, and 6x+y-\-58 = 0. 

4. Obtain the formula for the area of a triangle by drop- 
ping perpendiculars from each of the three vertices upon the 
X-axis, and considering the trapezoids formed. 

GENERAL PROBLEMS 

1. Show that the triangle whose vertices are (3, 2), 
(— 1, —3), and (—6, 1) is a right triangle. 

2. An isosceles right triangle is constructed with the hy- 
potenuse on the line x + 4 y = 10, and the vertex of the right 
angle at the point (3, 4). Find the coordinates of the other 
vertices. 

3. Find the equation of the line through the point (5, 6) 
which forms with the axes a triangle whose area is 80. Four 
solutions. 

4. Find the equation of a line through the point (— 1, 0) 
such that the given point bisects that portion of the line be- 
tween the axes. 

5. Find the equation of a line through the point (3, —.6) 
such that the given point divides that portion of the line be- 
tween the axes in the ratio 3 : — 1. 



Ch. IV, § 40] THE STRAIGHT LINE 61 

6. Find the equation of the line through the point (8, 2) 
such that the portion of it included between the lines x— 2y=6 
and x + y = 5 shall be (a) bisected at the point ; (6) equal 

Via. 

7. On the line y — 5 = a segment is laid off, having 
for the abscissas of its extremities 2 and 5, and upon this 
segment an equilateral triangle is constructed. What are the 
coordinates of the third vertex ? 

8. Find the point on the line 4?/ — 5# -{- 28 = which is 
equidistant from the points (1, 5) and (7, — 3). 

. 9. Find the points which are equidistant from the points 

i* (4, — 3) and (7, 1), and at a distance 3 from the line 15 x + Sy 
= 120. 

10. The coordinates of the vertices of a triangle are (5, 2), 
(4, — 7), and (3, 7). The side joining the first two points is 
divided in the ratio 4 : 7, and through this point lines are drawn 
parallel to the other sides. Find their points of intersection 
with the other sides. 

11. On each side of the triangle in problem 10, find the 
point which is equidistant from the other sides of the triangle. 

12. The equations of the sides of a complete quadrilateral 
are 2y+7oj=14, x— 2y=l, .u+4,y= — 4, and 7 a— 4?/=— 28. 
Show that the middle points of the three diagonals lie on a 
straight line. 

13. Show that the perpendiculars let fall from any point of 
the line 2 x + 11 y = 5 upon the two lines 24 x + 7 y = 20 and 
4 x — 3 y = 2 are equal to each other. 

14. Perpendiculars are dropped from the point (0, 4) to the 
sides of a triangle whose vertices are (1, o), (5, — 1), and 
(6, 0). Show that the feet of these perpendiculars lie on a line. 

15. Find the equation of a line through the intersection of 
the lines 2 x — 7 y = 3 and x -f 3 y = 8, which is perpendicular 
\o the line joining the origin to the intersection of these lines. 



62 ANALYTIC GEOMETRY [Ch. IV, § 40 

16. Show that the four points (2, 1), (5, 4), (4, 7), and (1, 4) 

are the vertices of a parallelogram. 

17. Find the area of the triangle formed by the three lines 
y = m^x + c 1} y = m& + c 2 , and x = 0. 

18. "What is the value of a if the three lines 3 x -+- y — 2 = 0, 
ax -j- 2 y — 3 = 0, and 2 x — y — 3 = meet in a point ? 

19. Lines are drawn through the vertices of a triangle 
parallel to the opposite sides of the triangle, and the intersec- 
tions of these lines are joined to the opposite vertices of the 
triangle. Show that the joining lines meet in a point. 

20. Prove analytically that the bisector of the interior angle 
of a triangle divides the opposite side into segments propor- 
tional to the adjacent sides of the triangle. 

21. Prove that all straight lines, for which - + - = -, pass 

a b o 
through a fixed point, and find the coordinates of that point. 



CHAPTER V 



POLAR COORDINATES 




41. In Art. 9, the polar system of coordinates was men- 
tioned. Let be a fixed point and OA a fixed line through 
it. Then the position of any point 
P is fixed if the angle A OP and 
the distance OP are given. The 
distance OP is called the radius 
vector of the point P and is rep- 
resented by p. Positive values 
of p are laid off from along the 
terminal line of the angle, negative values in the opposite 
direction. The angle A OP is called the vectorial angle 
and is represented by 0. The usual convention in regard 
to angles will be followed, — the anti-clockwise direction 

of rotation being con- 
sidered positive. These 
two quantities are called 
the polar coordinates of 
the point, and are written 
O, 6). The line OA is 
called the initial line, and 
the point 0, the origin 
or pole. 
It appears that, while any pair of coordinates determine 
a single point, there will be an indefinite number of pairs 
of coordinates which will give the same point ; for there 

63 




Fig. 40. 



64 ANALYTIC GEOMETRY [Ch. V, § 42 

will be an indefinite number of angles which have the 
same terminal line. If 6 is restricted to values between 
— 2 it and 2 7r, any point may be determined by four sets 
of coordinates. 

If the polar coordinates of the point P in Fig. 40 
are (p, 0), the same point may also be determined by 
(- p ,0 + 180°), (- />, - 180°), and (/o, 6 - 360°). 

PROBLEMS 

1. Plot the following points: (6, ^], (6, f tt), 

(- 10, -!«.), (- 2, 1), (4, 0), (- 5, 0), (0, »). 

2. Write polar coordinates of each point in problem 1, in 
which p and are both positive. 

3. Show that the distance between two points whose polar 
coordinates are (/>i,0i) and (p 2 , 2 ) is Vpi 2 -f/o 2 2 — 2/3 1/ o 2 cos(0 1 — 2 ). 

42. Equation of a locus. — If the law according to which 
a point moves is stated, it may often be translated into an 
equation connecting p and 6. For example, if a point is to 
remain at a distance a from the origin, the value of p for 
every such point is a, while 6 can vary at pleasure. The 
polar coordinate equation of a circle about the origin is, 
therefore, p = a. The equation of any line through the 
origin is evidently 6 = &, for on such lines the value of p 
is entirely unrestricted, while 6 is fixed. 

Again, as in Cartesian coordinates, an equation connect- 
ing p and 6 restricts the points which satisfy it to a series 
of positions which lie on some curve. The curve which 
contains all the points whose coordinates satisfy an equa- 
tion and no other points is called the locus of the equation, 
and the equation is spoken of as the equation of the locus. 



Ch. V, § 43] POLAR COORDINATES 65 

PROBLEMS 

1. What is the polar coordinate equation of a line which 
makes an angle of — with the initial line and which passes 
through the origin? 

2. What is the equation of a line parallel to the initial line 
and three units above it ? 

3. Show that the equation of any line in terms of a andp is 

p cos (6 — a) = p. 

4. Show that the equation of a circle of radius r about the 
point (pu 0j) is p 2 + /Oi 2 — 2 p x p cos (0 — $ x ) = y* 2 . 

5. A circle of radius r has its centre on the initial line and 
passes through the origin ; show that its equation is 

p = 2 r cos 0. 

43. Plotting in polar coordinates. — The method of 
finding the curve which is the locus of any equation in 
polar coordinates is similar to that employed in rectangu- 
lar coordinates. Sometimes the law of formation may be 
determined directly from the equation, but it is usually 
necessary to find various points on the curve by giving 
values to one of the coordinates and finding the corre- 
sponding values of the other ; these points are then 
plotted and a smooth curve drawn through them. 
Coordinate paper may be made by drawing circles about 
the origin at a unit's distance from each other, and lines 
through the origin, making any convenient angle with 
each other. On this the position of the points may be 
fixed accurately without measurement. 

For convenience in plotting we insert a table of the 
natural values of the trigonometric functions for every 5° 
from 0° to 90°. 



66 



ANALYTIC GEOMETRY 



[Ch. V, § 44 



44. Natural values of the sines, cosines, tangents, and 
cotangents. 



e 


sin 


cos 


tan 6 


cot 




0° 


.0000 


1.0000 


.0000 




90° 


5° 


.0872 


.9962 


.0875 


11.430 


85° 


10° 


.1736 


.9848 


.1763 


5.671 


80° 


15° 


.2588 


.9659 


.2679 


3.732 


75° 


20° 


.3420 


.9397 


.3640 


2.747 


70° 


25° 


.4226 


.9063 


.4663 


2.145 


65° 


30° 


.5000 


.8660 


.5774 


1.732 


60° 


35° 


.5736 


.8192 


.7002 


1.428 


55° 


40° 


.6428 


.7660 


.8391 


1.192 


50° 


45° 


.7071 


.7071 


1.0000 


1.000 


45° 




cos 


sine 


cot 6 


tan0 


e 



PROBLEMS 

1. Plot the locus of the equation p sin = 5. 

By the aid of the table of sines given above, the following 
points are seen to lie on the locus: (29, 10°), (14.6, 20°), 
(10, 30°), (7.8, 40°), (6.5, 50°), (5.8, 60°), (5.% 70°), (5.08, 80°), 
(5, 90°). 

Plotting these points on the coordinate paper, the locus 
appears to be a straight line MX, parallel to the initial line 
and five units above it. 

2. Plot the locus of the equation p 2 = 100 cos 2 0. 

If we let vary from 0° to 45°, the following points will be 
found to lie on the locus : (10, 0°), (9.7, 10°), (8.7, 20°), (7, 30°), 
(4, 40°), (0, 45°). There will also be the points (- 9.7, 10°), 
etc. The values of p will be imaginary for values of 6 between 
45° and 135°, and there will therefore be no real points corre- 
sponding to these values of 0. If we let $ vary from 135° to 
180°, the following points will be found on the locus : (0, 135°), 
(4, 140°), (7, 150°), (8.7, 160°), (9.7, 170°), (10, 180°). Also 



Ch. V, § 44] 



POLAR COORDINATES 



67 



(- 4, 140°), etc. If we let 6 vary from 180° to 360°, p will 
pass through the same changes in value as before, and the 
same points will be located. The curve has the form shown 




in Fig. 41. The two tangents to the curve at the origin 
make angles of 45° and 135° with the initial line, and are 
therefore perpendicular to each other. The curve is called the 
lemniscate. 

3. Plot the locus of each of the following equations : 



(a) p sin = a. 

(b) /O (l-cos0) = 2a. 

(c) P = 2a(l-cos0). 

(d) p 2 = a 2 sin 2 0. 

(e) p cos = a cos 2 6. 
(/) p = a cos 3 6. 

(g) p = a sin 4 6. 

(h) p 2 cos0 = a 2 sm30. 



(Q p 2 = a- cos 3 0. 

(J) p = a (sec -f tan 6). 

(k) p = a (cos 2 6 -f- sin 2 1 

(?) p = 2 a tan • sin 0. 

(m) P = a(l + 2cos0). 

(?i) p — aO. 

/^\ a 



CHAPTER VI 



TRANSFORMATION OF COORDINATES 



45. When the equation of a curve, referred to any sys- 
tem of coordinates, is known, it is often desirable to obtain 
the equation of the same curve, referred to some other 
system. If we know its equation in Cartesian coordinates, 
we may wish to obtain its equation in polar coordinates, 
or the reverse. Or, knowing its Cartesian equation re- 
ferred to a certain set of axes, we may wish to obtain its 
equation referred to some other set of axes, in order to 
obtain the simplest, or most useful form of its equation. 
This can be done, if we can obtain the relation connecting 
the coordinates of any point on the curve in the first sys- 
tem of coordinates and the coordinates of the same point 
in the second system. 

AVe can transform from any Cartesian system to any 

other by first changing 
the origin without chang- 
ing the direction of the 
axes, and then revolving 
each of the axes through 
some angle. 

46. Transformation to 

axes parallel to the original 

axes. — Let OX and OY 

F 1G . 42. be any given pair of rec- 



r 



-x 



M 



Ch. VI, § 46] TRANSFORMATION OF COORDINATES 69 

tangular axes, and let O'X' and 0' Y' be a new pair, par- 
allel to the old, and having for their origin the point 0', 
whose coordinates with respect to the original axes are x 
and y . Let P be any point, and let its coordinates in 
the first system be (x, y) and in the second (V, y') 6 

From the figure, OM = OA + AM, 
and MP = AO' + NP. 

But OM = x, OA = x , AM= x\ 

MP = y, AO'=y , NP = y' . 

Substituting these values, we find 

X - X + X', 

and y = yo + y' L J 

as the equations connecting the old and new coordinates, 
and these equations will be found to hold wherever the 
point P is placed in the plane. 

If we have an equation, which expresses the law of 
movement of a point by giving the relation between its 
coordinates referred to the first pair of axes, the substi- 
tution of these values for x and y will give the relation 
which must exist between x' and y f -, the coordinates of 
the point referred to the second pair of axes, in order 
that the point may move in the same path. It must be 
understood that x' and y' are variables, like x and y. 
The primes are only used to distinguish the coordinates 
used in the two systems, and may be dropped after the 
substitution has been made. 

In the above demonstration no use has been made of the 
fact that the axes are rectangular. The same formulas 
will therefore hold for transforming from any set of 
oblique axes to any parallel set. 



70 



ANALYTIC GEOMETRY 



[Ch. VI, § 47 



PROBLEMS 

1. If the equation of a line referred to any given system of 
Cartesian coordinates is 3 x -f 4 y = 10, what is its equation 
referred to a parallel system, the coordinates of whose origin, 
referred to the original axes, are (— 2, 5) ? 

The formulas connecting the old coordinates of any point 
with the new are x — _2-\-x' 

y = 5 + y'. 
Substituting these in the equation of the line, we have 

3 (-2 + x') + 4(5 + 2/') = 10, 
or reducing and dropping primes, 

3 x + 4 y == — 4. 

Construct the two sets of axes and plot the locus of each of 
the equations, showing that the same line will be obtained in 
both cases. 

2. The equation of a line is 4 x — 3 y = 8. Find the equa- 
tion of the same line, referred to a set of axes, parallel to the 
old, through the point (2, — 5) as origin. 

Plot the locus with respect to both axes. 

47. Transformation 
from one set of rectangu- 
lar axes to another, hav- 
ing the same origin and 
making an angle 9 with 
the first set. — Let OX 
and OYhe the given set 
of rectangular axes, and 
OX' and OY' another 
set of rectangular axes 
making an angle with the first set. Let the coordinates 
of P with respect to the original axes be (x, y), and witli 




Eig. 43. 



Ch. VI, § 48] TRANSFORMATION OF COORDINATES 71 

respect to the new axes (x 1 ', y'~). Draw its ordinates, MP 
and NP, and the lines KN and LN parallel to OX and 
OY. The angle at P is evidently eqnal to 0. 

OM=x, MP = y, 0~N=x\ NP = y' . 
Then 0M= OL-KN. 

But OL = ON cos 6=x r cos 0, 

and KN = NP sin = y' sin d. 

Hence x = x' cos - y' sin 0. [21, a] 

In like manner 

MP = LN+KP 

= (Wsinfl + iVPcostf 
or y = a?' sin + t/' cos 0. [21, b~\ 

48. Transformation in which both the position of the 
origin and the direction of the axes are changed. — If it be 

required to change the position of the origin to the point 
(% #o)> anc ^ a ^ ^ ne same time to revolve the axes through 
the angle 0, the two operations may be performed sepa- 
rately, or we may combine the two previous formulas into 

the one set, , , . 

x = Xo -f- x' cos 9 - y' sin 0, 

y — y -f jc< sin + y' cos 0. I J 

PROBLEMS 

1. Transform the equation 3x + 7 y = 8 to a new set of 
axes parallel to the old set, and having the point (4, — 2) 
as origin. 

2. Show that the equation x 2 + y 2 = a 2 , referred to rectan- 
gular axes, will be unchanged by revolving the axes through 
any angle, keeping the origin fixed. 

3. Transform the equation x 2 — y 2 = 10, referred to rectan- 
gular axes, to axes bisecting the angle between the old axes. 



72 



ANALYTIC GEOMETRY [Ch. VI, §§ 49, 50 



4. Through what angle must the coordinate axes be turned, 
if in its new position the X-axis goes through the point (o, 7) ? 

5. Given the equation xr -f y- -f 8 x — 4?/ = 0. To what 
point must the origin be changed to cause the terms in x and y 
to disappear ? 

6. Given the equation 2y 2 + 2 xy -f .r 2 + 4 = 0, referred to 
rectangular axes. Through what angle must the axes be 
turned to cause the term in xy to disappear ? 

49. Transformation from 
any Cartesian system to any 
other Cartesian system, hav- 
ing the same origin. — In 
Fig. 44, OX and OY are 
the original axes, and a> is 
,the angle between them ; 
OX' and OY' are the new 
axes; OX' and OY' make 
angles and <j> with OX. 
Let the student show that 




sine 



sill co sin CO 



II 



/ Sin(o> - 8) 
sino> 



What do these formulas become when m = 90° 5 
a> = 90° and <£ = 0? 



[23] 



When 



50. Degree of an equation not changed by transformation 
of coordinates. — The degree of an equation cannot be 
changed by transformation from one system of Cartesian 
coordinates to any other. For we have seen that in each 
case we replace x and y by expressions of the first degree 



Ch. VI, § 51] TRANSFORMATION OF COORDINATES 



73 



in x' and y\ and that therefore the degree of the equation 
cannot be raised. Neither can it be lowered, for it would 
then be necessary to raise the degree in transforming 
back to the original axes, since we must obtain the origi- 
nal equation. 



51. Transformation from rectangular to polar coordi- 
nates. — Let it be required to find the equations of 
transformation for transforming from a given set of rec- 
tangular axes, OX and OY, to a polar system having 
as its origin and OX as 
its initial line. 

The relations between 
#, y, />, and 6 are seen at 
once from the triangle 

OMP; for sin 6 = ^, 

, n OM 0F 

and cos u = — — , or 



OC = p cos 6, 
y = p sin 6. 



M 



[24] 



Fig. 45. 



The formulas for transformation from polar to rectan- 
gular coordinates are easily seen from the same triangle 
to be 

p 2 = a;2 + y\ 

X 

It is not, however, generally necessary to use this 
second set of formulas, as the transformation from polar 
to rectangular coordinates can usually be made more 
easily by the aid of the first set. 



74 ANALYTIC GEOMETRY [Ch. VI, § 51 

PROBLEMS 

1. Obtain the polar equation of the curve whose rectangular 
equation is x 2 + y 2 = r. 

Substituting .i* = p cos 0, and y = p sin 0, we have 

p- cos 2 6 4- p 2 sin 2 = r 2 , 
or p 2 = r 2 , or p = r. 

This is then the polar equation of the curve whose rectangu- 
lar equation is x 2 + if = r 2 . 

2. Obtain the polar equations of the curves whose rectangu- 
lar equations are 

(a) a¥ 4- &V = a 2 6 2 , (e) (.x* 2 -f y 2 ) 3 = 4 dhhf, 

(6) y 2 = 2 mx, (/) k 2 + r + a* = 0, 

(c) a?-tf = a 2 , (g) y 2 = X , 



(d) (x 2 + tff = a 2 (x 2 - y 2 ), (1i) x 2 + y 2 + 2 ax = a Vx 2 + y 2 . 

3. Obtain the rectangular equation of the curve whose polar 
equation is p = a cos 6. 

We might make this transformation by using the two for- 
mulas [25], but it will be found to be easier first to multiply 
both members of the equation by p, giving 
p- = a.p cos 0. 

Using the formulas x = p cos and p 2 = x 2 -+- y 2 , this reduces 
at once to x 2 + y 2 = ax. 

This is then the rectangular equation of the curve whose 
polar equation is p = a cos 6. 

4. Obtain the rectangular equations of the curves whose 
polar equations are 

(a) p = a sin 6, (/) p = a sin 2 6, 

(b) P = a + -A_, (g) p 2 cos 20 = a 2 , 

(c) p = a (1 + cos 6), (h) p = a (cos 2 6 + sin 2 0), 

(d) p 2 = a 2 cos 2 0. (i) p = a (1 + cos 2 0), 

(e) p = a — b cos 0, (./) p = 2 a tan • sin 0. 



CHAPTER VII 



THE CIRCLE 




52. Equation. — The locus of points equidistant from 
any fixed point is called a circle. Hence, to find the 
equation of a circle, it 
is necessary to express 
the algebraic relation be- 
tween the coordinates of 
such points. 

If the origin is taken 
at the centre of the circle, 

the equation is evidently 

x2 + V 1 — r2 i where r is 
the radius of the circle. 
For the distance of any 
point (#, y) from the origin is x 2 -f- y 2 . 

If the centre be taken at any point (7, whose coordinates 

are («, /3), the distance CP from the centre to any variable 

point P is V(# — a) 2 + (y — /3) 2 . Hence the equation of 

the circle is _ „ 

(aj-a)2 + (3f-02)=r2. [26] 

If the centre is on the X-axis, y3 = 0, and the equation 

reduces to 

(x — ct) 2 + y 2 = r-; 

if on the Y-axis, a = 0, and the equation reduces to 

x 2 +(y- /3) 2 = r 2 . 



Fig. 46. 



76 ANALYTIC GEOMETRY [Ch. VII, § 53 

Problem. — Find the equation of a circle, (a) tangent to 
both axes ; (b) passing through the origin and having its centre 
on the X-axis. 

53. General form of the equation. — Expanding [26J, 
we have 

x 2 + y i _ 2 ax - 2 (By + a 2 + /3 2 - r 2 = 0. 

a and /3 can have any value, positive or negative, and r 
can have any positive value. Hence the equation is in 
the general form of 

x 2 + y 2 + Doc + Ey + F= 0, [27] 

where D = - 2 a, U=-2/3, and JP = « 2 + /3 2 - r 2 . And 
if an equation is to represent a circle, it must be in the 
form of [27]. It will be noted that this is not the most 
general form of the equation of the second degree. For 

Ax 2 + Bxy + Cij 2 + Dx + Fy + F=0. 

When the two equations are compared, it will be seen 
that the term in xy is wanting in [27], and that the coeffi- 
cients of x 2 and y 2 are equal, or 

£=0, and A = C. 

Hence both these conditions must be satisfied in order 
that the general equation of the second degree may repre- 
sent a circle. 

But will it always represent a circle when these condi- 
tions are satisfied ? It will be necessary to determine 
whether there are always values of «, /3, and r which cor- 
respond to all values of D, F, and F. Solving the equa- 
tions given above for «, /3, and r, we have 

« = — , /3 = -=-^, and r = i r\/W+E 2 -±F. 



Ch. VII, § 54] THE CIRCLE 77 

Hence there will always be real values for a and ft for 
all values of D, E, and F. But if D 2 + E 2 - 4 F< 0, the 
value of r is imaginary, and there will be no point in the 
plane which will satisfy the equation. But since it has 
the form of the equation of a circle, it is said to represent 
an imaginary circle. 

Again, if D 2 + E 2 - 4 F= 0, r = 0, and the equation 
represents the point («, ft) only. It is called a null circle. 

We see then that we shall have a real circle only in 
case D 2 4- E 2 — 4 F > 0. But no equation in the form 
of [27] can represent any other locus. Hence it is said 
to represent a circle, 

real, if D 2 + E 2 - 4 F > ; 

null, if D2 + E 2 -±F=Q; 

imaginary, if D 2 + E 2 - 4 F < 0. 

54. Circle through three points. — We know from plane 
geometry that three points not in a straight line determine 
a circle. It ought therefore to be possible to find the 
equation of the circle passing through three such points, 
(x v 3/j), (> 2 , y 2 ), and 3 , y 3 ). This may be done by 
determining D, E, and F of the general equation [27] 
so that these coordinates will satisfy that equation. 

Substituting these coordinates successively in equation 
[27], we have 

• z 1 2 + !J 1 2 + I)x 1 + Ey 1 + F=0, 
x 2 2 + y 2 2 + Dx 2 + Ey 2 + F=0, 
x 2 + y s 2 + JDz 3 + E I/3 + F=0. 

From ^hese three equations it is always possible to 
determine D, E, and F (if the three points do not lie on a 



78 ANALYTIC GEOMETRY [Ch. VII, § 54 

line), and their values substituted in the general equation 
[27] will give the equation of the circle through the 
points. 

PROBLEMS 

1. What is the equation of a circle, if 

(a) its centre is at the point (—2, 3), and r = 6, 

(b) its centre is at the point (—3, —4), and r = 5, 

(c) its centre is at the point (5, 3), and it is tangent to 
the line 3a — 2y = 10, 

(d) its radius is 10, and it is tangent to the line 4 x-\-3y=70 
at the point (10, 10), 

(e) it passes through the three points (4, 0), ( — 2, 5), (0, —3), 
(/) it circumscribes the triangle, the equations of whose 

sides are x -\-2y — 5 = 0, 2x-\-y — 7 = 0, and x — y -f 1 = 0, 
(g) it has the line joining the points (3, 4) and (— 2, 0) as a 

diameter, 

(h) it passes through the points (5, — 3) and (0, 6) and has 

its centre on the line 2x — 3y = 6, 

(i) it passes through the points (5, —3) and (0, 6) and r=6? 

2. Find the coordinates of the centre and the radius of each 
of the following circles : 

(a) x 2 + y 2 + 8x-6y-10 = 0, 

(6) x 2 + y 2 + 8x-6y + 50 = 0, 

(c) x 2 + y 2 + 6y-16 = 0, 

(d) 3x 2 + 3y 2 -7x-8 = 0. 

3. Show that if the equations of two circles differ only in 
the constant term, they represent concentric circles. 

4. Show that the equation of a circle in oblique coordinates 
is in the form of 

x 2 + 2 cos to • xy 4- f + Dx + Ey 4- F = 0. 

What conditions must be satisfied by the general equation 
of the second degree that it may represent a circle when referred 
to any particular set of oblique coordinates ? 



Ch. VII, § 55] THE CIRCLE 79 

5. Show that the equation of any circle through the points 
of intersection of two given circles, 

x 2 +if + D Y x -f E$ + F 1 = 0, 

and a 2 + y 2 + A&* + E 2 y + F 2 = 0, 

can be expressed in the form 
x 2 + / + As + ^y + *i + * (s 2 + 2/ 2 + A* + A*/ + Ft) = 0. 
What is the locus of this equation when k = — 1 ? 

6. Obtain the equation of the common chord of the two 
circles, 

aP + tf + Sx — y = 0, 

and x 2 + ?/ 2 — 4 ?/ + 10 = 0, 

and show that it is perpendicular to their line of centres. 

7. Prove that the common chord of any pair of intersecting 
circles is perpendicular to their line of centres. 

8. What would be the statement of problems 5 and 7, if the 
two circles do not intersect ? 

55. Tangent. — A tangent to any curve is defined as 
follows : Let a secant through a fixed point P 1 of the 
curve intersect the curve again at P v Let P 2 move 
along the curve toward P v The secant will revolve 
about P v and as P 2 approaches P x the secant will ap- 
proach a certain limiting position. This line, which is 
the limiting position approached by the secant as P 2 ap- 
proaches P v is called the tangent to the curve at P v 

The method of finding the equation of the tangent to 
any curve of the second degree is the same for all. The 
demonstration should, therefore, be studied carefully in 
the case of the circle where the work is the simplest. 



80 



ANALYTIC GEOMETRY 



[Ch. VII, § 55 



According to the definition we must first write the 
equation of a secant through two points, and then find 
the limiting form which this equation approaches when 
the two points approach coincidence. 

Let (x v y^) and (x 1 + A, y x + &) be the coordinates 
of P 1 and P 2 , adjacent points on the circle x 2 -f- y 2 = r 2 . 
The equation of the line through these two points is 
(by [5]) 

y - ?/i = & 

x — x 1 h 

If we let P 2 approach P v h and k will approach zero, 
and the limit of the second member will be indeterminate. 

This would be neces- 
sary since w r e have 
made no use of the 
fact that P 2 must ap- 
proach P x along the 
circle. Unless P 2 ap- 
proaches P 1 along 
some curve, P X P 2 will 
have no limiting posi- 
tion. It will there- 
fore be necessary to 
determine in the case of each curve the value of the 

1c 

expression -• In the case of the circle about the origin, 

the coordinates of the points P 1 and P 2 must satisfy the 
equation x 2 + y 2 = r 2 . 

We have, therefore, (1) x^ + y^ = r 2 , 




Fig. 47. 



and (2) x 2 + 2 hx 1 + h 2 + yf + 2 ky x + Jc 2 = r 2 . 



Ch. VII, § 50 J THE CIRCLE 81 

Subtracting (1) from (2), we have 

2 hx x + h 2 + 2 %! + k 2 = 0, 

k 
or, transposing and solving for -, 

- — — %%{-{- h 
h~ 2y x -\-k 

Substituting in the former equation of the secant P^Py 

we see that y _ ?/i _ ^ 2x^+h 

x — x x 2 y 1 -\- k 

is another form of its equation in the circle x 2 + y 2 = r 2 . 

If now we let h and k decrease, the limit of the second 

member is no longer indeterminate, but becomes K 

The equation of the tangent is therefore Jl 

V ~ ffi = -% 
x - x 1 yl 

which by the aid of (1) reduces to 

asi* + yiy = r 2 . [28] 

Let the student show by the same method that the 

equation of the tangent to the circle 

a .2 + y i + j) x j r E y j r F=0 

is wi& + yiV + ^(.(B + i*h) + ^(y + yi)+F=0* [29] 

56. Normal. — The normal at any point of a curve is 
the line through the point, perpendicular to the tangent at 
the point. Its equation can be obtained by first writing 
the equation of the tangent at the point, and then that of 
a perpendicular to it through the point of contact. 
■ The equation of the normal to the circle x 2 + y 2 = ?* 2 , at 
the point (^x-^y^). is seen to be y x x — x Y y = 0. 



82 ANALYTIC GEOMETRY [Ch. VII, § 57 

PROBLEMS 

1. Obtain the equations of the tangents and normals to the 
following circles, and show that in each case the normal passes 
through the centre of the circle : 

(a) x* + y 2 = 25, at (3, 4), 

(b) x * + tf + 2x-4y + 5 f = 0, at (- 1, 2), 

(c) x 2 -f y 2 — 14 x — 4 y — o = 0, at the points whose abscissas 
are 10. 

(d) x 2 -f- y 2 — 6 x — 14 y — 3 = 0, at the points whose abscissas 
are 9. 

2. Find the angle in which the two circles x 2 + y 2 — 4 x = 1 
and a,* 2 -f y 2 — 2 y = 9 intersect. 

Note. — The angle between two curves is the angle between their 
tangents at the point of intersection. 

3. Show that the following circles cut each other orthogo- 
nally (or intersect at right angles) : 

x* + tf- Sx 4-4?/+ 7 = 0, 

a; 2 + y 2 - 10 x - 6 y + 21 = 0. 

4. Show that the length of the tangent from the point 
(pit Vi) to tne circle xr -f- y 2 -f Dx + Ey + F =0 is 



V.1-! 2 -f 2/i* + Dx^ 4- ^2/i + ^« 

Note. — Use the right triangle having for its legs the tangent and the 
radius to the point of contact. The length of the hypotenuse is the dis- 
tance from the point (&i, y{) to the centre of the circle. 

5. What is the length of the tangent from the point 
(— 2, 6) to the circle x 2 + y 2 4- 2 y = 5 ? 

57. Tangents from an exterior point. — The equation of 
the tangent which we have obtained can be applied only 
when we know the coordinates of the point of contact 



Ch. VII, § 57] 



THE CIRCLE 



83 



(x v y^). There are other conditions which will determine 
the tangent. Consider first the tangent from a given 
exterior point. The method of procedure may here be 
best shown by an illustration. 

Let it be required to find the equation of a tangent 
from the point (5, 10) to the circle whose equation is 

^2 + 3/2=100. 

Let the coordinates of the unknown point of contact 
be (x v y^). Then the equation of the tangent will be 
x x x + y x y = 100. Now this tangent is to pass through 
the point (5, 10), and therefore these coordinates must 
satisfy its equation, or 

5^ + 10^ = 100. 

This is one equation connecting x x and y v and the fact 
that the point (x v y^) lies on the circle gives another, 

:100. 



h 2 + Vi ■■ 



The algebraic solution of these equations gives 

x x =0, x 1 = 8, 

or 
^ = 10, ^ = 6. 

There are, therefore, as we should expect, two points of 
contact of tangents from the given exterior point, viz. : (0, 10) 
and (8, 6). Substituting these 
values in the equation of the 
tangent, we have 

10?, = 100, 

and 8 x + 6 y = 100, 

as the equations of the tangents 
through the point (5, 10). 




84 ANALYTIC GEOMETRY [Ch. VII, § 58 

PROBLEMS 

1. Obtain the equations of the tangents to the following 
circles : 

(a) x 2 + f = 49, from (6, 8). 

(b) x 2 + y 2 - 4 x -22 = 0, from (- 2, 6). 

(c) x 2 + y- + 5 y = 25, from (7, - 1). 

2. Obtain in each of the problems the equation of the line 
joining the points of contact of the two tangents. 

3. Obtain in this way the equation of the chord of contact 
of tangents from the exterior point (x^, y x ) to the circle 
x 2 -\- y 2 = ?- 2 . 

58. Tangent in terms of its slope. — When the slope of 
the tangent is given, we might proceed as in Art. 57, for 
we could obtain one equation by placing the slope of the 

tangent, -, equal to the given slope. Solving this 

with x^ + y x 2 = 100, we could find x x and y x just as before. 
But another method is more important. The equation of 
any line which has the given slope I may be written in the 
form 7 , , 

y=lx + b. 

It is then only necessary to find what value of b will 
make it a tangent to the circle x 2 + y 2 = r 2 . Every line 
of the system will cut the circle in two points, real, 
imaginary, or coincident. If the points are coincident, 
the line is a tangent. Starting the solution of y = lx + b 
and x 2 + y 2 = r 2 , we have at once by substitution 

(1 + p) a? + 2 Ibx + b 2 - r 2 = 

for determining the abscissas of the points of inter- 
section. 



Ch. VII, § 59] THE CIRCLE 85 

This will in general have two distinct roots, but (by 

Art. 8) if 

(2^) 2 =4(l + / 2 )(6 2 -r 2 ), 

these roots are equal. This equation therefore gives the 
value of 5, which makes the line a tangent. Solving for 
6, Ave have 

b = ± rVT+P. 

There are then two tangents to the circle which have 
any given slope. Their equations are 

y = lx± rVTT¥. [30] 



PROBLEMS 

1. Obtain the equations of the tangents to the circle 
x 2 4- y 2 = 49, which are (a) parallel to the line 3 x — 2 y = 10 ; 
(b) perpendicular to the same line. 

2. Obtain the equations of the tangents to the circle 
e 2 + y 2 . -f- 6 x = 0, which are perpendicular to the line 
a-3y + 4 = 0. 

3. Determine the relation between a, b, and r if the line 

-+^=1 is tangent to the circle x 2 + y 2 = r ^. 
a b 

4. Determine the value of k if the line 3 x — 4 y = k is 
tangent to the circle x 2 4- y 2 — 8 x -f- 12 y — 44 = 0. 

5. Find the condition which must be satisfied if the line 
Ax + By -f C = is tangent to the circle 

x 2 + y 2 + Dx -f Ey -f- F = 0. 

59. Chord of contact. — We have seen that, from any 
point P x outside the circle x 2 -f- y l = r 2 , two tangents can 
be drawn to the circle. Let it be required to find the 



86 



ANALYTIC GEOMETRY 



[On. VII, § 59 



equation of the chord P 2 P% through the two points of 
contact of these tangents. 

The equations of the tangents P 2 P\ and P S P 1 are 
t ^ . (V [28]) 



v + y& = *** 

and x 3 x + y 2 y = r 2 . 

Both these equations 
must be satisfied by 

Oi, #i). 

Hence x 2 x x + y 2 y x = r\ 
and x z x x + y^y x — r 2 . 

But these are just the 
conditions which must be 
satisfied if the points P 2 and P z are on the line 

ocix + V\V = r 2 . [31] 

This is, therefore, the equation of the line P 2 P 3 which 

is the chord of contact. 

It Avill be noted that this equation has the same form as 

the equation of the tangent. It represents the tangent if 

the point P 1 is on the circle; but if P 1 is outside the circle, 

it is the equation of the chord of contact. 

Let the student show that the equation of the chord of 

contact of tangents from an exterior point to the circle 
x 2 + y 2 + Dx + Ey + F = 




ocxoc + piij + — (as - 



xi)+f(v + yi)+F=o. 



[32] 



PROBLEMS 

1. Find the length of the chord of contact of tangents from 
the point (3, 4) to the circle x 2 + y 2 = 4. 



Ch. VII, § 59] THE CIRCLE 87 

2. Find the equation of the circle which touches the line 
2 x — y = 10 at the point (3, — 4) and passes through the 
point (5, 1). 

3. Find the equation of the circle which passes through the 
point (1, 1) and also through the intersections of the circles 
x-+y 2 — 8 x+±y=10, and x 2 +y 2 =5x. [See prob. 5, page 79.] 

4. Find the equations of the three common chords of the 
three circles in problem 1, page 84, and show that they inter- 
sect in a point. 

5. Find the equation of the circle inscribed in the triangle 
whose sides are represented by the equations 

4 x -f- 3 y = 10, x — 5 y = 15, and 3 x — 4 y = 8. 

6. Find the area of the triangle formed by the axes of 
coordinates and the tangent to the circle x 2 + y 2 = r 2 at the 
point (a-!, ft). 

7. Construct the circles x 2 + y 2 = x -f- 2 y, and x 2 + y 2 = 2 x. 
Find the equations of their line of centres, their common chord, 
and points of intersection. Show that their common chord is 
perpendicular to their line of centres. At what angles do the 
circles intersect ? 

8. Show that in any circle a line perpendicular to the tan- 
gent at the point of contact passes through the centre. 

9. Show that an angle inscribed in a semicircle is a right 
angle. 

10. Show that the perpendicular from any point of a circle 
on a chord is a mean proportional between the perpendiculars 
from the same point on the tangents at the extremities of the 
chord. 

11. If from any point on a circle circumscribed about a 
triangle perpendiculars are dropped to the sides of the triangle, 
the feet of these perpendiculars lie on a line. 

12. Show that the chord of contact of tangents from an 
exterior point is perpendicular to the line joining that point to 
the centre of the circle. 



CHAPTER YIII 

LOCI 

60. We have seen that when a property common to all 
points of a locus is given, the translation of this property 
into an algebraic equation between the coordinates of the 
points gives the equation of the locus ; for this is just what 
is meant by the equation of a locus, — an equation which is 
satisfied by the coordinates of every point which satisfies 
the given conditions, and by no other points. The actual 
work then always consists in this translation of a condition 
expressed in language into a relation between the coordinates 
expressed in an algebraic equation. Any method which 
enables us to do this may be employed. The simple 
methods have already been exemplified in the previous 
chapters. In these cases the law may be expressed as 
an equation in x and y at once by the aid only of a sim- 
ple geometrical construction. There are many problems 
which may be solved in this way. 

PROBLEMS 

1. The sum of the squares of the distances of a moving 
point from two fixed points is constant. Find the locus 
of the moving point. 

Let the X-axis pass through the fixed points, with the 
origin midway between them. Then (a, 0) and (— a, 0) 
will represent the points. Let (.r. y) be any position of 



Ch. VIII, § 60] 



LOCI 



89 




the moving point. Then placing the sum of the squares 
of the two distances equal to a constant, k< we have 

This is then the equation which 
must be satisfied by the coordi- 
nates of all points fulfilling the 
given conditions, and is there- 
fore the equation of the locus 
desired. Reducing, we have 

x 2 + y 2 = -- a 2 . 

This is the equation of a circle about the origin. This 
property of a circle might be used to define it as well as 
the more familiar one. Most curves may be defined in 
many ways ; for any property which is sufficient to 
determine the curve completely may be used as its 
definition. 

2. The difference of the squares of the distances of a 
moving point from two fixed points is constant. Show 
that its locus is a line perpendicular to the line through 
the two fixed points. 

3. The distances of a moving point from two fixed 
points are equal. Find the locus. 

4. The distances of a moving point from two fixed 
points are in the ratio of m to n. Show that the locus is 
a circle. Find the centre and radius, and show that, if 
m = 7i, the locus becomes the same as that obtained in 
the previous problem. 



90 ANALYTIC GEOMETRY [Ch. VIII, § 60 

5. Find the locus of the centres of circles of radius r, 
which pass through a fixed point (x v y x ). 

6. Find tlie locus of the centres of circles which pass 
through two fixed points. 

7. Find the locus of the centres of circles which touch 
two given lines. 

8. The sum of the squares of the distances of a moving 
point from the sides of a square is constant. Find the 
locus. 

9. Find the locus of a point, the square of whose dis- 
tance from a fixed point is m times its distances from a 
fixed line. 

Note. — Take the fixed line as the Y-axis, and let the X-axis pass 
through the fixed point. 

10. The sum of the squares of the distances of a moving 
point from the four corners of a fixed square is constant. 
Show that the locus is a circle whose centre is at the centre 
of the square. 

11. Given the base of a triangle and the distance from 
one end of the base to the middle point of the opposite 
side, find the locus of the vertex. 

12. The sum of the squares of the perpendiculars let 
fall from a moving point on the sides of an equilateral 
triangle is constant. Find its locus. 

13. The sum of the squares of the distances of a mov- 
ing point from r fixed points is constant. Show that its 
locus is a circle. 

14. A line of given length moves so that its ends shall 
always touch two lines at right angles to each other. 
Find the locus of the middle point. 



Ch. VIII, § 61] LOCI 91 

15. The three points (9, M, N lie on a line. Find the 
locus of the point P, when Z0PM=ZMPW. 

16. One side of a triangle and the angle opposite are 
fixed. Find the locus of the vertex of the angle. 

61. It is, however, often impossible to obtain easily 
by direct geometrical methods the relation between the 
coordinates. We may then find it necessary to introduce 
certain other auxiliary variables, which we call param- 
eters. These must be so chosen that it is possible to 
express in equations the relation between the coordinates, 
x and ^, of the moving point and these parameters. If 
we have introduced n parameters and can find n + 1 
independent equations, it is always possible to combine 
them in such a way as to eliminate all the n parameters 
and leave a single equation connecting the coordinates of 
the moving point. This resulting equation must be the 
equation of the locus. The difficulty of the elimination 
evidently increases with the number of parameters used. 
When more than two or three are used, it becomes very 
laborious. Care should therefore be taken to choose that 
method which requires the introduction of the fewest 
parameters. 

The following problems illustrate some of the methods. 

17. A straight line is drawn parallel to the base of 
a triangle and its extremities are joined tranversely to 
those of the base. Find the locus of the intersection 
of the joining lines. 

Choose the base of the triangle as the X-axis and a 
perpendicular to this side through the opposite vertex 



92 



ANALYTIC GEOMETRY 



[Ch. VIII, § 61 



as the F-axis. Let DE be any position of the line par- 
allel to the base. We are to find the locus of P\ the 
point of intersection of the lines 
CE and AD. Let the coordinates 
of A be (a, 0); of B, (0, b); of 
x (7, O, 0); and of P', (V, #')• 



C (C,o) 



B (o,l)) 




Y' 

Fig. 51. 



A (a,o) Any particular values of x' and y 
evidently depend upon the posi- 
tion of DE, and this depends upon 
a single parameter, its distance 
from CA. Let the equation of DE in any particular 
position be y = k. We need the equations of the lines 
CE and AD. It is therefore necessary to determine 
the coordinates of D and E. The equation of AB is 

- + f = 1, and by solving this equation with the equa- 
a o 

tion y = Jc, the coordinates of E are easily found to be 
[ — ^ — p — -, k). In like manner the coordinates of D are 

found to be l c ^ ~ \ k). 

The equations of AD and CE are 

&&e + [(# — c)5 -f £&];?/ — &#5, 

and &for -j- [(e — «)& -f- a¥]y = &<?&. 

Now since these two lines both pass through P\ its 
coordinates (V, y') must satisfy both equations, or 

kbx' -f- [(a — c)b -f <?&]?/ = &a&, 

and &fa/ + [(c — a)b + a¥\y } = kcb. 

Here then are two equations between x\ y\ and k. 
The elimination of k will give a single equation in x' 



Ch. VIII, § 61] LOCI 93 

and y' which must be the equation of the locus of P f . 

For it will be the algebraic expression of the relation 

which must exist between the coordinates of P\ that 

it may be the intersection of the two diagonals. The 

elimination is here easily performed. For, adding, we 

have 

2 kbx' + k(a + c)y' = kb(a + c). 

Dividing by k and dropping the primes, we have as 
the equation of the locus, 

2bx + (a + c)y = b(a + <?). 

Let the student find from the conditions of the problem 
two points through which the curve must pass, and test 
the result obtained above by substituting in it the coor- 
dinates of these points. 

18. Find the locus of the intersection of the diagonals 
of rectangles inscribed in a given triangle. 

19. On the sides of a given triangle measure off equal 
distances from the extremities of the base, and at these 
points erect perpendiculars to the sides. Find the locus 
of the point of intersection of these perpendiculars. 

20. The ends of the hypotenuse of a given right tri- 
angle touch the coordinate axes. Find the locus of the 
vertex of the right angle. 

21. Parallel lines are drawn with their ends on the 
two axes. Find the locus of the point which divides 
them in the ratio of m: n. 

22. One side and the opposite angle of a triangle are 
fixed. Find the locus of the centre of the inscribed 
circle. 



94 ANALYTIC GEOMETRY [Ce. VIII, § 61 

23. Each radius of the circle, x 2 + y 2 =r 2 , is extended 
a distance equal to the ordinate of its extremity. Find 
the locus of its terminal point. 

24. In a rectangle, ABCD, let EF and GrH be drawn 
parallel respectively to AB and BO. Find the locus of 
the intersection of HF and EG. 

25. In the previous problem let ABCD be any paral- 
lelogram and solve with the aid of oblique coordinates. 

26. Find the locus of the middle point of a system of 
parallel chords of the circle x 2 4- y 2 = r 2 . 

Let y — Ix -f- b be the equation of any one of the parallel 
chords ; let (x v y^) and (x v y 2 ) be the coordinates of 
the points where it cuts the circle, and (V, y'~) the coor- 
dinates of the point midway between these points. It 
is required to find an equation connecting x' and y' which 
may contain I but must not contain b. 

Starting the solution of the two equations, y = lx + b 
and x 2 + y 2 = r 2 , we have 

(1 + £2)^2 + o ibx 4- b 2 - r 2 = 0, 

the two roots of which must be x l and x 2 . 

But x 1 = ^ t X% ' 



Hence (1) x' = - ~y (See Art. 8.) 

Since the point (V, y') lies on the line y = lx + b, its 
coordinates must satisfy that equation, or 

(2) y' = lz'+b. 



Ch. VIII, § 61] 



LOCI 



95 



We have then two equations in x' , y', I, and 6, from 
which b must be eliminated. From (2) b = y' — lx'. Sub- 
stituting this value in (1) and 
reducing, we have as the equa- 
tion of the desired locus 
x + ly = 0. 

Since this equation is of the 
first degree and contains no 
constant term, it represents 
a straight line through the 
centre of the circle, and con- 
forms to the ordinary definition of a diameter, 
evidently perpendicular to the parallel chords. 




It 



27. Find the locus of the middle points of chords which 
pass through a fixed point Qx v y^) of the circle x 2 + y 2 = r 2 . 

Let P', (V, ?/'), be the middle point of any chord through 
P v (x v y^). Let (x v y 2 ~) be the coordinates of P 2 , the 
other extremity of the chord. From the formulas for 

bisecting a line [4], we have 




(1) *': 



X* -\~ Xn 



and (2) y' = U±±]h. 

And, since P 2 is a point on 
the circle, 

(3) x* + y*=r*. 
Here are three equations 
between the variables x' and y\ the constants x v y v and r, 
and the parameters x. 2 and y 2 . It is therefore possible 



96 



ANALYTIC GEOMETRY 



[Ch. VIII, § 61 



to eliminate the parameters and obtain a single equation 
in terms of the variables and constants only. Solving (1) 
and (2) for x 2 and y v we have 

x 2 = 2 x' - x v and y 2 = 2 y' - y v 

Substituting these values in (3), we have 

4 x' 2 + 4 y' 2 — 4 x x x' — 4 y x y' + x x 2 + y x 2 = r 2 . 

But x x 2 -j- y x 2 = r 2 , and, dropping primes, the equation 

reduces to 

x 2 + y 2 -x 1 x-y 1 y = 0. ■ 

This is the equation of the locus of P' . It is a circle on 

OP i as a diameter, since its centre is at the point [ -i, ^ 
and it passes through the origin. 

When, as in the above problem, we have to determine 
the locus of a point situated on a moving line which 
revolves about some fixed point in it, polar coordinates 
are often convenient. The fixed point is taken as the 
pole, and the distance from it to any position of the 

moving point becomes 
— A the radius vector. 

The following prob- 
lem will illustrate the 
method : 

28. Find the locus of 
the middle points of 
chords of the circle, 

x 2 + y 2 = r 2 , 

which pass through a fixed point, (x v y±), not on the 
circle. 




Cn. VIII, § 61] LOCI 97 

Let jPj be a fixed point through which the secant P X P 3 
passes, and let it be required to find the locus of P\ the 
middle point of P^P Z - Transform the equation of the 
circle to polar coordinates, with P 1 as origin. The equa- 
tions of transformation are (by [20] and [24]), 

x z= x, + p cos 0, 

(1) 

y = y x + p sin 0, 

and the equation of the circle becomes 

(2) p 2 + 2 (x x cos + y x sin 0) p + x t 2 + y? - r 2 = 0. 
Let p' and 0' be the polar coordinates of P' . The vec- 
torial angles of P v P 1 , and P 3 are evidently the same, 
and if 6' be substituted for in (2), the solution of the 
resulting equation, 

p 2 + 2 (>! cos 0' + ^ sin 0') /> + x 2 + y x 2 - r 2 = 0, 

for p will give p 2 and /o 3 , the two values of p for the points 
P 2 and P 3 . 



But 



, ?2 + / 



r _ 2 - 

and /> 2 -f- /> 3 = — 2(2^ cos0 r + ?/ x sin0'). (See Art. 8.) 

ence // = — (x x cos 0' + y x sin 0'). 

This equation expresses the relation which must exist 

between the polar coordinates of P\ and, dropping primes, 

we have as the polar equation of the locus, referred to P x 

as origin, 

p = — x x cos — y 1 sin 0. 

The equations for transforming back to rectangular 
coordinates, obtained from (1), are 

p cos = x — x v 
and p sin — y — y v 



98 ANALYTIC GEOMETRY [Ch. VIII, § 61 

From these we see that 

P 2 = O ~ ^i) 2 + Cy - ^i) 2 - 

If the polar equation of the locus is multiplied by />, 
and these values substituted, it becomes 

O - x i) 2 + (y - yO 2 = - v + *i 2 - y\y + y 2 , 

or x 2 + y 2 — x x x — y x y — 0. 

This is the rectangular equation of the locus referred to 
the original origin, and is seen to represent a circle on 
OP x as diameter. 

29. Solve problem 27 by means of polar coordinates, 
and problem 28 by means of rectangular coordinates. 

30. Find the locus of the points which divide in the 
ratio m : n chords through a fixed point (x v y^) of the 
circle x 2 + y 2 = r 2 . 

31. Lines through a fixed point P x cut the circle 
x 2 -f y 2 = r 2 in the points P 2 and P 3 . Find the locus of a 
point P of this line, if 

P P — 2 ^1^3 X ^1-^3 . 

PiPi + PiP* 

32. Chords through a fixed point of a circle are extended 
their own length. Find the locus of their extremity. 

33. Lines are drawn from a fixed point P v meeting a 
fixed circle in P 2 . On P X P 2 a point P is taken so that 
P X P x P^ 2 = k 2 . Find the locus of P. 

34. Lines are drawn from a fixed point P v meeting 
a fixed line in P 2 . Find the locus of the point which 
divides P X P 2 1J1 the ratio m : n. 



Ch. VIII, § 61] LOCI 99 

35. Lines are drawn from a fixed point P v meeting a 
fixed line in P v Find the locus of a point P on these 
lines if P X P x P X P 2 = k 2 . 

36. Find the locus of points from which tangents to 
two given fixed circles are equal. (See problem 4, page 
82.) Show that the locus is a line perpendicular, to the 
line joining the centres of the two circles. 



CHAPTER IX 



CONIC SECTIONS 

62. Definition and equation. — If « a point moves so that 
the numerical ratio of its distance from a fixed point to its 

distance from a fixed line 
remains constant, its locus 
is called a conic. 

Let the fixed line be 
taken as the y-axis, and 
a perpendicular through 
the fixed point F as the 
X-axis. Let the perpen- 
dicular distance OF of 
the fixed point from the 
fixed line be represented by m. Let P be any position 
of the moving point. Then we are to find the equation 
of the locus of P when 
FP 
MP 




Fig. 55. 



(i) 



= (any constant) = e. 



But 



FP = ^(x-m) 2 +y 2 , and MP = x. 



Then (1) becomes ^ — m ^ + ^ 



or 
or 



x 2 — 2 mx -f- m 2 -f- y 2 = e 2 x 2 , 

(1 - e 2 ) ac 2 - 2 nix + y' 1 4- m 2 = 0. 
100 



[33] 



Ch. IX, § 63] CONIC SECTIONS 101 

This is then the general equation of a conic. The form 
which it takes in any particular case depends upon the 
values given to the constants, m and e. 

The fixed line OM is called the directrix of the conic, 
and the fixed point F, the focus. The value of the con- 
stant ratio e is called the eccentricity. The line perpen- 
dicular to the directrix through the focus is called the 
transverse or principal axis of the conic. The points where 
the transverse axis cuts the conic are the vertices. 

63. Parabola. e = l. 

When e = 1, equation [33] reduces to 

y 2 — 2 mx + m 2 = 0. 

This curve has but one vertex, which is evidently midway 
between and F. For when y = 0, x = — -. The equa- 
tion will be reduced to a simpler form, if we transform to 
this point as origin. The equations of transformation are 
(by [20]) 

x = x + — , and y = y' . 



Substituting these values, the equation becomes 

y 2 = 2 irioc. [3^] 

From this equation we see that the curve passes through 
the origin ; that it is symmetrical with respect to the 
X-axis ; that it is real only to the right of the y-axis ; 
and that as x increases, y increases, — ■ at first more 
rapidly than x, until x—— , then more and more slowly. 
It has, therefore, the form shown in Fig. 56. 



102 



ANALYTIC GEOMETRY 



[Ch. IX , § 63 



But, for the study of the distant points, polar coordinates 
ire bettei adapted. Transforming to polar coordinates 

with as ori 
ZI\ Y 




equa 



tion 



>4] becomes 
2 m cos 



Eig. 5G. 



sin 2 tf 

When == 0, p is infinite, 
and the curve, therefore, 
does not cut the X-axis 
a second time. But if 
we give to 6 any finite 
value, however small, p 
will have a finite value, 
which will be very large 
for small values cf 0, and will decrease as 6 increases, 
until for 6 = — , p — 0. We see, then, that every line 

through except the X-axis cuts the curve a second time, a 
fact which does not appear from the rectangular form of 
the equation. Yet the discussion of that form showed 
that the curve constantly recedes from the X-axis. It can 
be shown by the aid of the equation of the tangent that 
tlte curve approaches parallelism with the X-axis. 

TIiis particular species of conic is called the parabola. 

We have already defined the line MN as the directrix ; 
the point F as the focus ; OX as the principal axis ; and 
as the vertex cf the curve. We saw that 

DO = \T>F=\m. 
The coordinates of the focus, referred to as origin, are 
therefore ( — , ). The equation of the directrix is ;r= — . 



Ch. IX, § 64] CONIC SECTIONS ■ 103 

The line LR through F, perpendicular to the X-axis 
and terminated by the curve, is called the latus rectum. 

The abscissa of R is seen to be — , and by substituting this 

value for x in the equation of the parabola, its ordinate is 
found to be m. The length of the latus rectum is there- 
fore 2 m. 

PROBLEMS 

1. What is the equation of the parabola having its vertex 
at the origin, and its focus (a) on the X-axis, at a distance — 

to the left of the origin ; (&) on the ]T-axis, above the origin ; 
(c) on the I^-axis, below the origin ? 

2. What is the equation of the parabola if the focus is at 

the origin and the vertex at a distance — to the left of the 
origin ? 

3. What is the equation of the parabola, if its vertex is at 
the point («, /3) and its axis is parallel to the X-axis ? 

4. What is the equation of the parabola which has its ver- 
tex at the origin and passes through the points (3, — 4) and 
(-3, -4)? 

5. Obtain the equation of the directrix, the coordinates of 
the focus, and the length of the latus rectum in the parabola 

64. Central conies. e=£l. 

We see from the form of the equation of a conic, 

(1 - e 2 > 2 - 2 mx + f + m 2 = 0, [33] 

that it always represents a curve symmetrical with respect 
to the X-axis. When e = 1, we have seen that there is 
but one value of x for each value of y. But when e =£ 1, 
there will be, in general, two numerically unequal values 



104 



ANALYTIC GEOMETRY 



[Ch. IX, § 64 



of x for any given value of y. The curve is therefore 
not symmetrical with respect to the Z"-axis. But it will 
be shown to be symmetrical with respect to a line parallel 

to that axis. Transform 
the equation to a new ori- 
gin midway between the 
points where the curve 
1 X cuts the X-axis. The 
F"-axis will then be found 
to be an axis of symmetry. 
Placing */ = in [33], 
we have 



O A' 



Fig. 57. 



(1 — e 2 )x 2 — 2 mx + m 2 = 0. 

The two solutions of this equation will give the inter- 
cepts, OA and OA', on the X-axis. Let these be denoted 
by x x and x 2 . But we wish to know 00 (=#), C being 
the middle point of A' A. 

Hence « x 



2 



But we know that the sum of the roots of a quadratic 

is , where a and b are the coefficients of x 2 and x 

a 

respectively. (Art. 8.) 



Hence 



X l + X 2 = - 



-, and x - 



The equations for transforming from to as origin 
will then be (by [20]) 



1- 



-, and y = y' . 



Ch. IX, § 64] CONIC SECTIONS 105 

Substituting in [33], it becomes 



,„ 2 mx f 



(1 - e 2 ) 2 _ 
Reducing and dropping primes, 



Dividing by 



x 2 



■ + -?— = 1. 

.9... 9 — x * 



Let =-== # 2 * and the equation becomes 

(1 - e 2 ) 2 H 

^ + 2,/ a , = L [ 35 J 

a 2 a J (l — e 2 ) 

If then this transformation is possible, we have reduced 
[33] to a form which represents a curve symmetrical with 
respect to both axes. It is always possible except for 
the case when e — 1. But if e = 1, no value could be 
obtained for x , and the point O would not exist. This 
has been discussed in Art. 63. All other cases are in- 
cluded in equation [35]. 

The intercepts of the curve on the new axes, obtained 
from equation [35], are ± a and ± aVl — e 2 . This equa- 
tion must therefore represent two classes of curves quite 
dissimilar in form ; for while all intercepts are real when 
e < 1, we see that the intercepts on the Praxis will be 
imaginary when e > 1. If, when e < 1, we let the inter- 



106 ANALYTIC GEOMETRY [Ch. IX, § 65 



cepts on the F-axis, ± aVl — e 2 , be represented by ± 5, 
equation [35] becomes 

But since we wish to work with equations having only 
real coefficients, b cannot represent the same expression 
when e > 1, for Vl — e 2 would be imaginary. We then 
let ± aVe 2 — 1 = ± £', and equation [35] becomes 

S-S- 1 - [^ 

The b used in the first case is the actual intercept. 
In the second case b' is the real coefficient of the imagi- 
nary intercept, and 

b 2 = - V 2 . 

We see then that there are three distinct forms which 
the locus may take. If e = 1, the conic has been called a 
parabola ; if e < 1, it is called an ellipse ; and if e > 1, an 
hyperbola. The ellipse and hyperbola are called central 
conies to distinguish them from the non-central conic, the 
parabola. They may be treated together from the single 
equation [35], or from their separate equations. 

Let the student show that, if the directrix is taken 
as the X-axis, and a perpendicular to it through the 
focus as the ]T-axis, the simplest equation of the central 

conies is — — — + '2_=1. What is its form for the 

a 2 (l — e 2 ) a 2 

ellipse ? hyperbola ? 

65. Ellipse. e < 1. 

If, in Art. 64, we had solved the equation 
(1 — £ 2 ) x 2 — 2 mx + m 2 = 0, 



Ch. IX. § 65] 



CONIC SECTIONS 



107 



we would have found for the intercepts on the X-axis, 
and 



m 

! I+7 



i-i 



When e < 1, both these intercepts are positive, and there- 
fore both A and A! lie to the right of the directrix OY. 
One intercept, x v is evi- 
dently less than m, and ^ Y' 
therefore one point of 
intersection is between 
and F'. Since OC= - 



O A T 



-, OC>OF', and the 



points must take the po- 
sitions indicated on the 

X JWi. OO. 

figure. 

We have shown that, when e < 1, the equation of the 
conic referred to the new axes (see Fig. 58) is 



£ y 



p 



1. 



[36] 



From the form of this equation we see that the curve is 
symmetrical with respect to both axes, and hence to their 
intersection ; that it cuts the X-axis at the points 
(± a, 0), and the y-axis at (0, ± b) ; that the values of x 
are real only for values of y from — b to + b ; and that 
the values of y are real only for values of x from — a to 
-h a. A more careful plotting of the points will show 
that it has the form shown in Fig. 59. 

The line D'H' has been called the directrix, and the 
point F f the focus. Place the points F and D on the 
X-axis so that CF= F'Q and CD = D f C, and draw DH 
perpendicular to the X-axis. The symmetry of the curve 



108 



ANALYTIC GEOMETRY 



[Ch. IX, § G5 



shows that if we had used the line DH and the point F as 
directrix and focus, and the same value of e, the same 
curve would have been found as the locus. The curve 
can be said therefore to have these two lines DR and 
D'H' as directrices, and the two points F and F' as foci. 




We can now obtain the relations between the various 
lines in the figure. We have seen that 

FD = D'F' = m, 



OD = 


= B'C 


= 




m 


1 


-e 2 


CA-- 


= A'0 


= 


a 


em 


1-e 2 


nn - 


- wr? 




l 


em 



It follows that CD = -, and that the equations of the 
directrices are 



x = % and sc = -^L 
e e 



[38] 



Ch. IX, § 65) 



CONIC SECTIONS 



109 



Also that OF=CD-FD = -^— - m = -tS!L. = ae. 

1 — e 1 1 — e 2 

It is convenient to let OF be represented by a single 
letter c. 



Then 



c = ae, or e = - 



In obtaining equation [36], we let b 2 = a 2 (l — e 2 ). 
Solving for e 2 , we have 

e , = q^J>* t |- 39 -j 



Comparing these two values of e, we have 
a 2 — 6 2 = c 2 . 



[40] 



From this we see that BF, being the hypotenuse of a 
right triangle whose legs are c and 6, is equal to a. It 
also shows that a is 
always larger than 5, or 
that A'A(=2a), the 
axis perpendicular to the 
directrices, is larger than 
B'B(=2b). A' A has 
been called the trans- 
verse or principal axis ; 
B'B is called the conju- 
gate or minor axis of the 
curve. 

If the foci of the ellipse 
are on the F"-axis, the 
vertex A also lies on that 

axis, and B on the X-axis (Fig. 60). Its equation is (see 
end of Art. 64) 




V 



+ ^=1. 



[41] 



110 ANALYTIC GEOMETRY [Ch. IX, § 66 

All the formulas found above hold for [41], except the 
equations of the directrices, which are 

»-*r 

PROBLEMS 

1. Find a, b, c, e, and the equations of the directrices in the 
ellipse, 

(a) 4a 2 + 9/ = 36, , (b) 9a 2 + 4/ = 36, 

(c) 3 ar + $ if = 10. 

2. Find the equation of the ellipse having its centre at the 
origin and its foci on the X-axis, if 

(a) a ■ = 3 and b = 2, (d) b = 4 and c = 3, 

(6) 6 = 3 and e = i (e) a = 5 and c = 3, 

(c) a = 6 and e = f, (/) c = 4 and e = -|. 

3. Show that the length of the latus rectum (line through 

2 b 2 
the focus perpendicular to the axis) of the ellipse is 

4. Show that the circle is the limiting form of the ellipse 
as a and b approach equality. What is the eccentricity of the 
circle, and where are its foci and directrices ? 

5. What is the equation of the ellipse which has its centre 
at the origin and its axes coincident with the coordinate axes, 
and which passes through the points (4, 1) and (— 3, 2) ? 

6. What is the equation of an ellipse if its centre is at the 
point (a, /3) and its axes are parallel to the coordinate axes ? 

66. Hyperbola. e>l. 

When e > 1, one of the intercepts, , is positive, 

1 + e 

and less than m, while the other, , is negative. OC 

will also be negative. The points, A, A', C, and JP, will, 
therefore, take the positions indicated in Fig. 61. 



Ch. IX, § < 



CONIC SECTIONS 



111 



We have shown that, when e > 1, the equation of the 

conic reduces to 

Y' 



x 2 



b' 2 ' 



1. [37] 



AF 



Fig. 61. 



Again we see that the 
curve is symmetrical with 
respect to both axes, and 
hence with respect to the 
origin; that it cuts the X-axis at the points (± a, 0), and 
does not cut the T"-axis ; and that the values of y are real 
only for values of x numerically equal to or greater than a. 
The exact form can be obtained more readily from the polar 

equation. Transforming — — ^— = 1 to polar coordinates 
a 2 

with C as origin, we have 



V 2 



/> 2 = 



a 2 b' 2 



b' 2 cos 2 d-a 2 sm 2 6' 



When 6 = 0, p = ± a, and as 6 increases, the denomi- 
nator decreases, the fraction increases, and the point 
recedes from the origin. This will continue as long as 
the denominator remains positive. As soon as the de- 
nominator becomes negative, the value of p becomes 
imaginary. There is then a value of 6 beyond which 
the curve does not exist. This value of 6 is that which 
makes the denominator, V 2 cos 2 6 — a 2 sin 2 0, zero, or 



= tan - 



■(4> 



For 

tan-l- 



value of 6 between tan _1 ( -f- 



and 



every 

there will be a real value of /o, these val- 



ues growing larger as 6 approaches tan _1 ( H — ) or 



112 



ANALYTIC GEOMETRY 



[Ch. IX, § ( 



tan _1 ( ). The lines then which pass through the 

V «/ f b'\ f V\ 
origin, making the angles tan l f -\ — J and tan -1 ( J 

with the X-axis, do not cut the curve, while every line 
lying between these lines cuts the curve in two real points. 
The curve must therefore approach parallelism with these 
lines as the point recedes from the origin, and it will be 
shown in the next article that the curve approaches coinci- 
dence with these lines. Such a line is called an asymptote. 
If we continue to increase 0, we see that there will be 
no real value of p until tan 6 again becomes numerically 

V 
less than — Then p goes through the same changes 

in value, decreasing until it equals ± a. But we have 
shown that the curve is symmetrical with respect to both 
axes, and there is therefore no need of discussion beyond the 
first quadrant. The following is the form of the hyperbola: 




Place the points F' and D' on the X-axis so that 
F'0=CF and D f C = CD, and draw D' H' perpendicular 



Ch. IX, § 66] CONIC SECTIONS 113 

to the .X-axis. The symmetry of the curve again shows, 
as in the ellipse, that the hyperbola may be said to have 
two foci, .Fand F\ and two directrices, DJ3"and B 1 W '. 

We can now obtain the relations between the various 
lines in the figure. We have seen that 

DF=FD ! = m, 

OB = B' O = (- 00, Fig. 61) = 



e*-l 



OA = A'C = - 



e 2 -l 



OB = B' = V = —^- 



Vr-l 

It follows that OB = -, and that the equations of the 
directrices are 

x = ^, and *=-"• [42] 

Also that 

CF=CD + DF=-^— + m = 



It is convenient to let OF be represented by a single 
letter c. 

Then c = ae, 

c 

or e = — • 

a 

In obtaining equation [37], we let b' 2 = a 2 (e 2 — 1). 

Solving for e 2 , we have 

+ m *±2* [43] 

Comparing the two values of e, we have 

a 2 + b' 2 = c 2 . [44] 



114 ANALYTIC GEOMETRY [Ch. IX, § 67 

There is, in the hyperbola, no restriction on the relative 
values of a and b' . 

Note. — In the following articles we shall follow the ordinary custom, 
and use b in place of b'. 

67. Asymptotes. — The slopes of the asymptotes were 
seen [Art. 66~] to be ± -. Hence their equations are 

b , b 
y = -x, and y — x ; 

or written as a single equation, 

^ 2 _y 2 ==0 

a 2 P 

They are evidently the diagonals of the rectangle formed 
by drawing lines parallel to the axes through A, A\ B, 
and B f . 

It remains to be shown that the curve not only approaches 
parallelism with these lines, but actually approaches coinci- 
dence with them ; or that the perpendicular distance P X M 
from any point P x on the hyperbola to the asymptote 
decreases indefinitely, as I > 1 recedes from the origin along 
the curve. (See Fig. 62.) 

Since the equation of the asymptote is bx — ay = 0, 

PM = bx i ~ a Vi . (By [17]) 

V6 2 + a 2 

But, since P x is a point on the curve, 

b 2 x 2 - a V = a%\ 
or, factoring, 

bx 1 - ay x : 



bx x . + ay x 



Ch. IX, § 67] CONIC SECTIONS 115 

Hence P X M= 



(bxi + ayd^/tf + a 2 

This expression evidently decreases as x x and y x in- 
crease, approaching zero as a limit. The carve therefore 
approaches its asymptote indefinitely. 

PROBLEMS 

1. Find a, b, c, e, and the equations of the directrices and 
asymptotes of the hyperbola, 

(a) x 1 - 25 f = 25, (6) 9 x~ -±y 2 = 36, (c) 2x 2 -5if = 20. 

2. Find the equation of the hyperbola having its centre at 
the origin and its foci on the X-axis, if 

(a) a = 3 and b = 2, (d) b = 4 and c = 5, 

(6) b = 3 and e = 2, (e) a == 4 and c = 5, 

(c) a = 5 and e = f , (/) c = 10 and e = 3. 

3. What is the equation of an hyperbola, if its centre is at 
the point («, (3) and its axes are parallel to the coordinate axes? 

4. What is the equation of the hyperbola which has its 
centre at the origin and its foci on the X-axis, and which 
passes through the points (5, 3) and (—3, 2) ? 

2 b- 

5. Show that the latus rectum of the hyperbola is 

6. Show that the foot of the perpendicular from the focus 
of an hyperbola on an asymptote is at the distance a from the 
centre and b from the focus. 

7. Show that the circle of radius b, whose centre is at the 
focus of an hyperbola, is tangent to the asymptote at the point 
where it is cut by the directrix. 

8. Show that the product of the two perpendiculars let fall 
from any point of an hyperbola on the asymptotes is constant. 



116 



ANALYTIC GEOMETRY 



[Ch. IX, § 68 



68- Conjugate hyperbolas. — If, in deriving the equa- 
tion of the conic, the directrix is taken as the X-axis, and 
a perpendicular to it through the focus as the P'-axis, its 




Fig. 63. 

simplest form, in the case of the hyperbola, is 

~2 



b 2 



+ *= = !■ 



If the definitions of a and b are interchanged, using b to 
represent the semi- transverse axis (which is here the real 
intercept of the hyperbola on the Y"-axis), the equation 
becomes 9 9 



b* 



The hyperbolas ^-fj=l and ^-p : 



: — 1, where a 



a 2 " b 2 " a 2 

and b have the same values, are closely related. The 

transverse and conjugate axes of the first are respectively 



Ch. IX, § 69] CONIC SECTIONS 117 

the conjugate and transverse axes of the second. Two 
hyperbolas which are so related are called conjugate 
hyperbolas, either being conjugate to the other. But it 
is convenient to speak of the first as the primary and the 
second as the conjugate hyperbola. 

The polar equation of the conjugate hyperbola.; 

2 = - am 

9 b 2 cos 2 6 - a 2 sin 2 

differs from that of the primary hyperbola only in the sign 
of the second member. It therefore gives real values for 
p only for those values of 6 which gave imaginary values 
for p in the primary hyperbolas. The conjugate hyperbola 
has therefore the same asymptotes as the primary, but 
is situated on the opposite sides of those asymptotes. 

The value of c(=Va 2 + b 2 ') is the same for both the 
primary and conjugate hyperbolas, and the four foci there- 
fore lie on a circle having its centre at the origin. But 
for the conjugate hyperbola e f = -, and the equations of 

the directrices are y = ± — • 

e' 

69. Equilateral or rectangular hyperbola. — If b = a, 
the equation of the hyperbola becomes x 2 — y 2 — a 2 . This 
is called the equilateral hyperbola. The equations of its 
asymptotes are x + y = and x — y = 0. They therefore 
make an angle of 45° with the X-axis, and are perpen- 
dicular to each other. From this fact it is often spoken 
of as the rectangular hyperbola. 

The equilateral hyperbola and its conjugate are evi- 
dently equal to each other. Figure 68 shows an equi- 
lateral hyperbola and its conjugate. 



118 



ANALYTIC GEOMETRY 



[Ch. IX. 



PROBLEMS 

1. Write the equation of an hyperbola conjugate to the 
hyperbola 4 :r — y- = 4. and find its axes, eccentricity, latus 
rectum, the coordinates of its foci, and the equations of its 
directrices. 

2. Find the eccentricity of the equilateral hyperbola. 

3. Show that if e and e' are the eccentricities of two conju- 
gate hyperbolas, - -\ — = 1 and ae = be'. 

e e' 

4. Find the equation of the equilateral hyperbola referred 
to its asymptotes as axes. 

Xote. — Revolve the axes through the angle — lo : . 

70. Focal radii of a central conic. — The distance of a 
point on a conic from a focus is called a focal radius of the 
■ point. Since there are two foci in a central conic, there 
will be two focal radii for each point of the conic. 

The distance FP X of 
any point (x v y Y ) of the 
ellipse from the right- 
hand focus (ae, 0) is 

(by [1]) 




VCrj - aef + y*. ■ 

But since (x v y x ) is a 
point on the ellipse 



Fig. 64. 



its coordinates must satisfy this equation. 



Hence 



Vi 



2 = £ 2 - 



Ch. IX, § 70] 



CONIC SECTIONS 



119 



Substituting this value of y-f in the expression for the 
distance, we have 



Noting that 
this reduces to 



FP^yJx*- 
a 2 -b 2 



a' 



2 aex l + a 2 e 2 + b 2 x?. 

a 2 

= e 2 , and a 2 e 2 + b 2 = a 2 , 



FP 1 = Va 2 - 2 aex x + e 2 x 2 = ± (a - ezf). 

The distance, F'P V of the point from the left-hand 
focus may be found in the same way except that the coor- 
dinates of F' are ( — ae, 0) . 

Hence F'P 1 = ± (a + ex{). 

Let the student show that, though the work in the case 
of the hyperbola will be slightly different, the results will 
be the same. 

Since it is only the length of the focal radii that we 
seek, it will be neces- 
sary to determine in 
each conic which sign 
should be used before 
the parenthesis, so that 
it may express a posi- 
tive distance. In the 
ellipse a is always 

greater than ex v and the positive sign must therefore be 
used in both cases. 

Hence, in the ellipse, 

FP 1 = a + ex u 
and F'P 1 == a - ex u 




[45] 



120 ANALYTIC GEOMETRY [Ch. IX, § 71 

It will be necessary to consider the two branches of the 
hyperbola separately. For the right-hand branch ex 1 is 
always positive and greater than a ; and the two dis- 
tances are 

FP X = e&i - a, 

[46, a] 
and F'P X = exi + a. 

For the left-hand branch ex 1 is negative and greater in 
absolute value than a ; and the two distances are 

FP, = — ex\ + a, 

[46, b] 
and F' 'P x = — exi - a. 

From these results we see that the sum of the two focal 
radii of any point of an ellipse is 2 a. While in the hyper- 
bola the difference of the two focal radii is 2 a. The 
ellipse might therefore be defined as the locus of points, 
the sum of whose distances from two fixed points is con- 
stant ; and the hyperbola as the locus of points, the 
difference of whose distances from two fixed points is 
constant. 

Let the student obtain the equations of the ellipse and 
hyperbola in their ordinary forms from these definitions. 

71. Mechanical construction of the conies. — The results 
of the last section enable us to construct mechanically the 
ellipse and hyperbola. 

To construct the ellipse fix two pins at the foci, and 
place around them a loop of string whose length is 
2 a + 2 c. If a pencil is placed in the loop and moved 
about the foci, keeping the string taut, it will describe an 
ellipse. The proof is evident. 



Ch. IX, § 72] 



CONIC SECTIONS 



121 




An hyperbola may be traced in the following manner : 
Fix one end of a ruler at 
one focus, F'. A string 
whose length is 2 a less than 
the length of the ruler is 
attached to the focus F and 
to the other end of the ruler. 
A pencil, which holds the 
string taut and against the ruler, will trace a branch of 
the hyperbola. For in all positions of P, 

F'P -FP=2a. 

The parabola is per- 
haps more easily traced 
by points. Erect a per- 
pendicular ML at any 
point of the axis. With 
Pas a centre and a radius 
equal to DM, describe an 
arc, cutting ML at P. 
Then P is a point of the 
parabola. For iVP = PP. 
As many points as we 

please may be found in this way and the parabola passed 

through them. 

72. Auxiliary circles. — The auxiliary circle of a cen- 
tral conic is a circle described on the major axis as 
diameter. Its equation is x 2 + y 2 = a 2 . 

The circle described on the minor axis as diameter 
is called the minor auxiliary circle. Its equation is 
x 2 + y 2 = b\ 











L 


N 






._ .^ 


P 








\ 








/ 








/ 








D 












F 


F 

/ 


M 

/ 



Fig. 67. 



122 



ANALYTIC GEOMETRY , 



[Ch. IX, § 72 



Points on the ellipse and auxiliary circle which have 
the same abscissa are called corresponding points. In 




Fig. 68. 

Fig. 68, P x and R are corresponding points. The angle 
M X CR is called the eccentric angle of the point P v 

There is a simple relation between the ordinates of the 
corresponding points P 1 and R which may be found in 
the following manner. Let the coordinates of P x be 
(x v y x ), and of R (x v y 2 ). 

Substituting these values for x and y in the equations 
of the ellipse and circle respectively, we have 

b 2 x^ + a 2 y^ = a 2 b 2 , 
and x x 2 + y 2 2 = a 2 . 

Multiplying the second equation by b 2 and subtracting, 

WeW a 2 y 2 = b 2 y 2 , 

Ui - ._ b 



Or, the ordinate of any point of the ellipse is to the ordinate 
of the corresponding point of the circle as b is to a. 



Ch. IX. § 73] CONIC SECTIONS 123 

Let the student show that a similar relation holds 
between the abscissas of points on the ellipse and minor 
auxiliary circle which have the same ordinate. These are 
also called corresponding points. Show that CR passes 
through the corresponding point in the minor auxiliary 
circle. 

PROBLEMS 

1 . Find the focal radii of the ellipse xr + 9 y 2 = 18 for the 
points whose abscissa is — 2. 

2. Find the focal radii of the hyperbola 9 x 2 — 4 y 2 = 65 for 
the points whose ordinate is 2. 

3. Show that the distance of a point on an equilateral 
hyperbola from the centre is a mean proportional between the 
focal radii of the point. 

4. Prove that the circle described on any focal radius of an 
ellipse as a diameter is tangent to the auxiliary circle. 

5. Prove that the auxiliary circle of an hyperbola passes 
through the intersections of the directrices and asymptotes. 

6. Show that the focal radius of any point of a parabola is 

7. Show that the area of the ellipse is irab. 

Notk. — Divide the major axis of the ellipse into any number of equal 
parts, and on each of these parts inscribe rectangles in the ellipse and 
auxiliary circle. The areas of these rectangles will be in the ratio b : a, 
and by the theory of limits it may be shown that the areas of the ellipse 
and auxiliary circle will be in the same ratio. 

73. General equation of conies when axes are parallel to 
the coordinate axes. — From the equations of the conies 
which have been determined, we may obtain by transfor- 
mation of coordinates the most general form of their 
equations referred to any set of axes. A full discussion 



124 ANALYTIC GEOMETRY [Ch. IX, § 73 

of this question will be taken up in Chapter XIII., where 
it will be shown that every equation of the second degree 
represents some form of the conic. For the present we 
shall content ourselves with a very short discussion of 
their equations when the coordinate axes are parallel to 
the axes of the conic. When this is the case and the 
coordinates of the centre are («, /3), the equations of the 
central conies take the form (by [20]) 

(x-a) 2 . (y-/3Y = 1 
a 2 b 2 

The parabola whose vertex is at the point (a, /3), and 
whose axis is parallel to the X-axis, takes the form 

(y-{3) 2 = ±2m(x-a). 

If the axis of the parabola is parallel to the F"-axis, its 

equation is ^„ 

H 0-a) 2 =±2m(2/-/3). 

In each case the term in xy is wanting, and all of the 

equations are seen to be special cases of the general 

equation , „ „ „ _ ,, ,, 

4 Ax 2 + Cy 2 + Dx + Ey + F = 0. 

If neither A nor O is zero in this equation, it may be 
written in the form 

i (^ , $ ■ V 2 \ \ n ( o . E _. E 2 \ D 2 tf 2 w 

or, if we represent the second member by K, 
2>\ 2 / , E 



e + o) (» + m 



+ v ~ =i. 



K K 

A C 



Ch. IX, § 73] CONIC SECTIONS 125 

If A and C have the same sign, this takes the form of 
the equation of the ellipse whose centre is at the point 

, ], and in which a — \j — , and b =\ — . The 

2 A 20 J X A * C 

ellipse will be real, null, or imaginary, according as a and 
b are real, zero, or imaginary. 

Let the student show that if A and C have opposite 
signs, the equation represents an hyperbola, or (if K= 0) 
two intersecting lines. 

Also that, if either A or is zero, the equation repre- 
sents a parabola, or a pair of parallel lines. 

PROBLEMS 

1. Determine the nature and position of the locus of 

2 or 9 +3y 2 - 6x + ±y = 10. 
This equation may be written in the form 

2(^-3* + f) + 30/+J2/ + f) = 10 + f + f=^, 

i'2 TS 

The locus is an ellipse, having its centre at the point 
(1 3 — -§), and in which a = Vy4) an d b = V-f-f- 

2. Determine the nature and position of the locus of the 
following equations : 

(a) x? + 2y 2 -6x + y = 10, (d) Sx 2 - if + 6 y = 0, 

(b) x* + 4:X-2y = W, (e) y 2 + 2x-±y = 6 

(c) ±x 2 -3y 2 -±x + 8=0, 

3. Obtain the polar equation of each of the conies, the focus 
being used as the origin and the transverse axis as the initial 
line. 



CHAPTER X 



TANGENTS 

74. The method of finding the equation of a tangent to 
any conic at a given point is the same as that used in the 
case of the circle (Art. 55). The equation of a secant 
through the given point P v (x v y^), and an adjacent 

point P 2 , {x 1 + h,y l + k), 




on the curve is 

y - y\ , 



Fig. 



h 
h 

It is necessary to de- 
termine in each case a 

k 
value of - which will 
h 

not be indeterminate 
when h and k approach 

zero. We shall give the work in detail for the ellipse, 

b 2 x 2 + a 2 y 2 = a 2 b 2 . 

Since the points P x and P 2 lie on the ellipse, their 

coordinates must satisfy its equation, or 

(1) b 2 x 2 + a 2 y 2 = a 2 b\ 

(2) b 2 x 2 + 2 b 2 hx l + b 2 h 2 + a 2 y 2 + 2 a 2 hy x + a 2 Jc 2 = a 2 b 2 . 
Subtracting (1) from (2), we have 

2 b 2 hx ± + b 2 h 2 + 2 a 2 ky x + a 2 k 2 = 0, 

Jc 2 b 2 x, + b 2 h 

or — = ■ — L — ! — 

h 2 a 2 y x + a 2 k 

126 



Ch. X, § 74] TANGENTS 127 

The equation of the secant may therefore be written 

y-y, = __ 2 b 2 x , + m 
x — x 1 2 a 2 y l + a 2 k 

Now, if we let P 2 approach P v h and k will approach 

zero, and the limit of the second member is no longer 

b 2 x 
indeterminate, but becomes - 1 -- 

a V\ 
The equation of the tangent is therefore 



y-y^ = 



b 2 x, 



x-x x a 2 y x 

or clearing of fractions and transposing, 

b 2 x x x + a 2 yiy = b 2 x^ + a 2 y£. 
But b 2 x 2 + a 2 y 2 = a 2 b\ 

and the equation of the tangent reduces to 

Wxiac + aHjxy = a 2 b 2 . [47] 

Let the student show that the equation of the tangent to 
the hyperbola, 

b 2 x 2 — a 2 y 2 — a 2 b 2 , is b 2 xxx - aHj^y - a-b 2 , [48] 
the parabola, y 2 = 2mx, is yxy = m(x + xi), [^9] 

the locus of the general equation of the second degree, 
Ax 2 + Bxy + Of + Dx + Ey + F= 0, 

is Ax x x + y (x x y + y\X) + Cy x y 

[50] 
+ ^(x + nc 1 ) + ^{y + y 1 )+F=0. 

These formulas can be most easily remembered and 
applied if we notice that they may be obtained from the 



128 ANALYTIC GEOMETRY [Ch. X, § 75 

equation of the conic by replacing x 2 and y 2 by x x x and 

, x-,1/ + y-,x . . x -\- x-i , w+y, 

y x y ; xy by w ^ Ul ; x and y by 2 1 and ^ ; the 

constant quantities being unchanged. 

The method of finding the equations of the tangents 
from an exterior point is the same as that given for the 
circle. (See Art. 57.) 

Let the student show that the equation of the chord of 
contact of tangents from an exterior point will, in each 
case, take the same form as the equation of a tangent at 
the point of contact. (See Art. 59.) 

75. Normals. — The normal at any point of a conic is 
the line through the point, perpendicular to the tangent 
at the point. 

It can be found in any case by writing the equation of 
a tangent, and then writing the equation of a perpendicu- 
lar to the tangent through the point of contact. 

For example, the tangent to the ellipse has been found 
to be b 2 x x x + cPyiy = <fib 2 . A perpendicular to this line 
will have the form a 2 y x x — b 2 x^y = k. Since the normal 
passes through P v h = « 2 j/i a i — ^ 2x \Vv an( ^ ^ ne equation 
of the normal becomes 

a 2 y x x - b\v A y = (a 2 - b 2 ) Xl y v 

In like manner, the equation of the normal to the 

hyperbola is 

a 2 y x x + b 2 x x y = («- + b 2 )x x y v 

and to the parabola is 

y x x + my = x x y x + my v 

The student should note that these formulas apply only 
when the equations of the curves are in the simplest form. 



Ch. X, § 76] 



TANGENTS 



129 



He is advised not to use them in solving problems, as 
they are not easily remembered, but in each case to write 
the equation of the tangent and then that of a perpen- 
dicular to the tangent at the point of contact. 

76. Subtangents and subnormals. — The projections on 
the X-axis of those parts of the tangent and normal 
included * between the point of contact and the X-axis 
are called the subtangent and subnormal. 




Fig. 70. 



In the ellipse (Fig. 70) M X T is the subtangent and 
M X N is the subnormal for the point P v The equation 
of the tangent at P 1 is b 2 x x x -f a 2 y x y — a 2 b 2 , and the equa- 
tion of the normal is a 2 y x x — b 2 x x y = (a 2 — b 2 }x x y v Find- 
ing the intercepts on the X-axis, we have 



6T=-, and CN= ^ ~ ^ x v 



But M X T= CT- CM X = -- 



and M l N=CN-CM 1 = 



(a 2 - b 2 )x, 



[51] 
[52] 



130 ANALYTIC GEOMETRY [Ch. X, § 76 



Let the student show that for the hyperbola - — — = 1, 

a 2 b 2 

the subtangent equals a ~ Xl > [53] 

the subnormal equals —,xi 9 [54] 
and that for the parabola y 2 = 2 mx, 

the subtangent equals - 2 asi, [55] 

the subnormal equals m, L ,r >6] 

PROBLEMS 

1. Find the equations of the tangents and normals to 

(a) 3 a? + 4^ = 19 at (1, 2), 

(6) 2x 2 -y 2 = U at (3, -2), 

(c) / = 6 a- at (6, - 6), 

(d) 2 a 2 - 3 xy + 6 x - 2 = at (2, 3). 

2. Find the lengths of the subtangents and subnormals 
in (a), (b), and (c), problem 1. 

3. Eind the equations of the tangents to 

(a) 16^ + 25^ = 400 from (3,4), 

(b) f = 4x from (-3, -2), 

(c) ar o -32/ 2 +2* + 19 = from (-1,2). 

4. Find the chords of contact of the tangents in problem 3. 

5. Find the lengths of the tangents and normals in (a), (6), 
and (c), problem 1. 

Note. — The terms "length of tangent" and " length of normal" are 
used to indicate the distances on the tangent and normal from the point 
of contact to the points where they cut the X-axis. 

6. Find the angles between the ellipse 4 x 2 + y 2 = 5, and 
the parabola y 2 = 8 x, at their points of intersection. 

7. Show how the subtangent or subnormal in the parabola 
may be used to construct the tangent at any point of the curve. 



Ch. X, § 77] TANGENTS 131 

8. From the fact that the subnormal in the parabola is 
constant, show that the tangent approaches parallelism with 
the axis as the point of contact recedes from the origin. 

9. Show that if the normals of an ellipse pass through 
the centre, the ellipse is a circle. 

10. Show that the distance from the focus of a parabola 
to any tangent is one-half the length of the corresponding 
normal. 

11. Show that the focus of a parabola bisects the portion 
of the axis intercepted by a tangent and the corresponding 
normal. 

77. Slope form of the equations of tangents. — For many 
problems it is convenient to have the equation of the tan- 
gent in terms of its slope only. This can be found for 
each of the conies just as it was found for the circle in 
Art. 58. 

We shall give the outline of the work for the ellipse. 
Starting the solution of the equation of any line, y=lx + f5, 
with the equation of the ellipse, b 2 x 2 + a 2 y 2 = a 2 b 2 , we have, 
for obtaining the abscissas of the points of intersection, 
the equation 

(b 2 + a 2 l 2 ) x 2 + 2 a 2 l(3x + a 2 (/3 2 - b 2 ) = 0. 

There will be two solutions of this equation, and hence 

two points where the line cuts the ellipse. But if (see 

Art. 8) 

4 aH 2 f& = 4 (b 2 + a 2 l 2 ) (a 2 /^ - a 2 b 2 ), 



or j3 = ±^/a 2 P + b 2 , 

the two solutions of this equation are equal, the two points 
of intersection of the line with the ellipse have become 
coincident, and the line is tangent to the ellipse. 



132 ANALYTIC GEOMETRY [Ch. X, § 77 

The equation of a tangent having a given slope I is 

therefore 

y = loc± VaH 2 4- 6 2 . [57] 

Let the student show that the equation of the tangent 
having a given slope I is 

for the hyperbola, b 2 x 2 — a 2 y 2 = a 2 b 2 , 

y = lx± y/aW - b\ [58] 

for the parabola, y 2 = 2 mx, 

y^in+ft [59] 

PROBLEMS 

1. Find the equations of the tangents to the ellipse 
4 x 2 + y 2 = 4 which are parallel to the line 2 x — 4 y -f 5 = 0. 

2. Find the equation of the normal to the parabola y 2 = 8x, 
which is parallel to the line 2x -\-3y = 10. 

3. Find the equations of the tangents to the ellipse 
x 2 + 2 y 2 — x -f y = 0, which are perpendicular to the line 
x — 5 y = 6. 

4. Find the condition which must be satisfied if the line 

y = lx + B is tangent to the hyperbola ^- = — 1. 

a- 6- 

5. Find the points on each of the conies where the tangents 
are equally inclined to the axes. For what case is the solution 
impossible ? 

6. Find the points on the ellipse and hyperbola where the 
tangents are parallel to the line joining the positive extremi- 
ties of the axes. 

7. Show that the line - -f- «L = 1 is tangent to 

a ft 

(a) the ellipse £ + 1 = 1, if £ + | = 1 ; 

a 2 b~ a 2 (r 



or b z a 

(c) the parabola y 2 = 2 ma,', if ma + 2 /? 2 = 0. 



(6) the hyperbola - - V- = 1 if ^L _ * . = 1 ; 
W JF a 2 b 2 a 2 B 2 



Ch. X, § 78] TANGENTS 133 

8. Find the equations of the common tangents to the ellipse 
a? + 9 y 2 = 9, and the circle x 2 + y 2 = 4. 

Note. — Write the equation of the tangent to each curve in the slope 
form and determine the value of I which will make the two equations 
identical. 

9. Find the equations of the common tangents to the 
ellipse 4 x 2 -f 9 y 2 = 36, and the hyperbola x 2 — y 2 = 16. Show 
that there would be no common tangent, if the second member 
of the equation of the hyperbola had any value less than 9. 
Why ? Construct the figure. 

10. Find the equations of the common tangents to the circle 
x 2 -f- y 2 = 9, and the parabola y 2 = 8 x. How many solutions 
are there ? Why ? Construct the figure. 

11. Show that two tangents can be drawn to any conic from 
an exterior point. 

12. Through any given point how many normals can be 
drawn to (a) an ellipse, (b) a parabola ? 

13. Obtain the equation of the tangent at the point P 1 of 
the parabola y 2 = 2 mx, by determining I in the slope form of 
the equation of the tangent in terms of x^ and y v 

78. Theorems concerning tangents and normals. — 1. The 

tangent and normal at any point of an ellipse bisect the 
exterior and interior angles respectively between the focal 
radii drawn to the point of contact. 

Let P X T and P X N be the tangent and normal to the 

ellipse -— + &- = 1 at the point P v We wish to show that 
a 1 b l 

P X T bisects the angle FP X K, and that P X N bisects the 
angle F'P^F. 

It is a well-known theorem of elementary geometry that 
the bisector of an interior angle of a triangle divides the 
opposite side into segments which are proportional to the 



134 



ANALYTIC GEOMETRY 



[Ch. X, § 78 



adjacent sides of the triangle. The converse theorem is 
also true. 

It is therefore sufficient to show that 



F'N 
: NF' 



F'F, 
FP X 

The equation of the normal P X N is 

a 2 y x x - b 2 x x y = (a 2 - b 2 )x^ v 




Fig. 71. 

. . a 2 —b 2 
Its intercept ON on the X-axis is — x v or, since in 

a? — b 2 a 

the ellipse — = e 2 , 

a 1 

CN=e 2 x v 
Also, F f C=CF=ae. 

Hence F'N= F'C+CN=ae + e 2 x x = e (a + ex{), 
and NF=CF - CN = ae - e 2 x x = e(a - exj. 

But (by [45]) 

F'F 1 = a + g^, and FP X = a — e# r 



Ch. X, § 78] 



TANGENTS 



135 



F'P F'N 
Hence l = , and the normal bisects the angle 

F'P^F. Since the tangent is perpendicular to the normal, 
it bisects the supplementary angle FP X K. 

Note. — It is upon this principle that whispering galleries are con- 
structed. If the whole or part of the sides of a room is a surface formed 
by revolving an ellipse about its major axis, all waves of sound, light, or 
heat starting from one focus and striking this surface will be reflected to 
the other focus. 

2. In an hyperbola the tangent and normal at any point 
bisect the interior and exterior angles respectively between the 
focal radii. 

3. If an ellipse and hyperbola are confocal (or have the 
same foci), they intersect orthogonally (or at right angles). 




Fig. 72. 



Since the direction of a curve at any point is along the 
tangent at that point, two curves intersect orthogonally, 
if their tangents at the point of intersection are perpen- 
dicular to each other. This is evidently the case here, 
since the tangent to the ellipse bisects the exterior angle 



136 



ANALYTIC GEOMETRY 



[Ch.X, § 78 



between the focal radii, and the tangent to the hyperbola 
bisects the interior angle. The curves therefore intersect 
orthogonally. 

4. The tangent at any point of a parabola makes equal 
angles with the focal radius drawn to the point of contact, 
and with the axis of the curve. 

In the parabola y 2 = 2 mx, let P X T and P X N be the tan- 
gent and normal at P v Join P X P and draw P^K parallel 




Fig. 73. 

to OX. We wish to prove that the angles TP X P and 
FTP X are equal. 

If we let the abscissa of P x be x v its ordinate will be 
±V2mx v since P t is a point on the parabola. Then the 
equation of the tangent at P ± is 



± V2 mx 1 • y = m (x + x^). 
If in this equation we let y = 0, we find the intercept 
OT to be - x r 
Hence TO = x v and 17=^ + 3. 



But 



",=m 



+ 2 mx 1 = x 1 + - 



Ch. X, § 78 



TANGENTS 



1.3T 



Hence TF= FP V and the angles TP X F and FTP X are 

equal. What other angles are also equal in the figure ? 

Note. — Parabolic reflectors depend on this principle. If a surface is 
formed by revolving a parabola about its axis, all waves of light, etc., 
which start from the focus will be reflected in lines parallel to the axis of 
the parabola. 

5. Two parabolas which have the same focus and axis, 
but which are turned in opposite directions, cut each other 
orthogonally. 

6. The chord of contact of tangents to a parabola from 
any point on the directrix passes through the focus. 




Fig. 74. 



The coordinates of any point L on the directrix may 
be represented by ( — ^, y\ The equation of P\P V the 
chord of contact of tangents from this point, is 



y x y = mx - 



138 ANALYTIC GEOMETRY [Ch. X, § 78 

Since the coordinates of the focus (^0) satisfy this 

equation, the chord of contact must pass through the 
focus. 

Let the student prove the converse theorem, viz. : Tan- 
gents at the ends of any focal chord meet on the directrix. 

7. Prove that the same theorems hold for the ellipse 
and hyperbola. 

8. Any two perpendicular tangents to the parabola meet 
on the directrix. 

Two perpendicular tangents may be represented by the 
equations m 

j x ml 

and y = — - — — • 

9 I 2 

By solving these equations simultaneously, the point of 
intersection of the two tangents is found to be 



Vm m(l - Z 2 )~ | 
L 2' 21 J' 



which is evidently a point on the directrix. 

Let the student prove the converse theorem, viz. : Two 
tangents to a parabola from any point on the directrix are 
perpendicular to each other. 

9. Tangents to a parabola at the ends of any focal chord 
are perpendicular to each other. 

10. Show that theorems 8 and 9 do not hold for cen- 
tral conies, but that perpendicular tangents (a) to an 
ellipse meet on the circle x 2 + y 2 = a 2 + b 2 ; (6) to an hyper- 
bola meet on the circle x 2 + y 2 = a 2 — b 2 . (See Chap. 14, 
Prob. 1.) 

11. The line joining any point on the directrix of a pa- 



Ch. X, § 78] 



TANGENTS 



139 



rabola to the focus is perpendicular to the chord of contact 
of tangents from the point. 

Take the coordinates of the point L (Fig. 74) on the 

directrix as ( , y x ), and show that the line LF which 

joins this point to the focus is perpendicular to the chord 

of contact y x y — mx — — - 

A 

12. Prove the same theorem for the central conies. 

13. The two tangents which may he drawn from an 
exterior point to any conic subtend equal angles at the focus. 

14. In the parabola the perpendicular from the focus on 
any tangent meets it on the tangent at the vertex ; the per- 
pendicular meets the directrix on the line through the point 
parallel to the axis of the parabola. 

The equation of the 
tangent at any point (x v 

y x y = mx + mx v 
The equation of a per- 
pendicular to the tangent 
through the focus is 

my, 
y x x + my = -^l. 

The coordinates of the 
point of intersection of 
these two lines are 



<••§> 




Fig. 75. 



They therefore meet on the l^-axis, which is the tan- 
gent at the vertex. 

Let the student prove the second part of the theorem. 



140 ANALYTIC GEOMETRY [Ch. X, § 78 

15. Show that theorem 14 does not hold for central 
conies, but that the perpendiculars from the foci of a cen- 
tral conic on any tangent meet the tangent on the circle 
x 2 + f = a 2 m ( See c hap> 14? Prob# 5^ 

16. The perpendicular from a focus on any tangent to 
a central conic meets the corresponding directrix on the line 
joining the centre to the point of contact of the tangent. 

17. In any conic, tangents at the ends of the latus rectum 
meet the X-axis on the directrix. 

18. The tangent at any point of the parabola meets the 
directrix and latus rectum produced at points equally dis- 
tant from the focus. 

19. The product of the perpendiculars from the foci 
of a central conic on any tangent is constant and equal 
to b\ 

20. The semi-minor axis b of a central conic is a mean 
proportional between a normal and the distance from the 
centre to the corresponding tangent. 

21. The tangents at any point of an ellipse and the cor- 
responding point on the auxiliary circle pass through the 
same point on the axis. 

PROBLEMS 

1. Show that, if the point of contact of a tangent to an 
hyperbola moves off along the curve, the tangent approaches 
the asymptote as its limiting position. 

2. Find the equations of the tangents to the hyperbola 
which pass through the centre. (Use slope form of the equa- 
tion of the tangent.) 

3. Show that the portion of any tangent to an hyperbola 
included between the asymptotes is bisected at the point of 
contact. 



Ch. X, § 78] TANGENTS 141 

4 Show that the area of the triangle formed by any tan- 
gent to an hyperbola and the asymptotes is constant. 

5. Through any point of an hyperbola parallels to the 
asymptotes are drawn. Show that the area of the parallelo- 
gram formed by these lines and the asymptotes is constant. 

6. Show that the normal at one extremity of the latus 
rectum of the parabola is parallel to the tangent at the other 
extremity of the latus rectum. 

7. Obtain the equation of the parabola referred to tangents 
at the ends of the latus rectum as coordinate axes. 

8. Show that the distances of the vertex and focus of a 
parabola from the tangent at one end of the latus rectum are 
in the ratio of 1 : 2. 

9. Show that the directrix of a parabola is tangent to the 
circle described on any focal chord as a diameter. 

10. Show that the tangent at the vertex of a parabola is 
tangent to the circle described on any focal radius of the 
parabola as a diameter. 

11. The tangent and normal at a point of the ellipse form 
an isosceles triangle with the X-axis. Find the coordinates of 
the point. 

12. Prove that the angle between two tangents to a parabola 
is one-half of the angle between the focal chords drawn to the 
points of contact. 

13. Show that the length of a normal in an equilateral 
hyperbola is equal to the distance of the point of contact from 
the centre. 

14. Find the points on the conjugate axis of an hyperbola 
from which tangents to the hyperbola are perpendicular to 
each other. 



CHAPTER XI 



DIAMETERS 



79. The diameter of a conic may be denned as the locus 
of the middle points of a system of parallel chords. The 
method of finding this locus has already been illustrated 
for the circle on page 94. 



We shall repeat the work for the ellipse 



x 2 






Let y = l x x + /3 be the equation of any one of the parallel 
chords ; let (x v y^) and (x v y^) be the coordinates of the 




Fig. 76. 



points where it cuts the circle, and (V, y f ~) the coordinates 
of the point midway between these two points. 

142 



Ch. XI, § 79] 



DIAMETERS 



143 



Starting the solution of the two equations, y = l 1 x + ft 
and b 2 x 2 + a 2 y 2 = a 2 b 2 * we have 

Q> 2 + a\ 2 ) x 2 + 2 a%#r + a 2 (ft 2 - I 2 ) = 0, 

the two roots of which must be x x and x 2 . 



But 

Hence 



a\ft 



[By Art. 8] 



b 2 + fl 2 ^ 2 

Since (V, «/') is on the line y = l x x -f & y may be found 
by substituting the value of x' in that equation. This 
gives 

y'=- m 



Hence 



b 2 + a\* 

b 2 ' 



or, dropping primes, we have as the equation of the 

diameter, 

Wx + a 2 hy = 0. [60] 

Y 




Fig. 78. 



144 ANALYTIC GEOMETRY [Ch. XI, § 80 

Let the student show that the equation of a diameter oi 

the hyperbola, —^ - &- = 1, is Woe - a*hy = 0, [61] 

the parabola, y 2 = 2 mx, is hy = m. [62] 

TAe /orm of the equation of a diameter of an ellipse or 
hyperbola shows that it passes through the centre of the conic, 
and that it therefore conforms to the ordinary definition 
of a diameter. But all the diameters of a parabola are 
seen to be parallel to the axis. 

Since any value may be given to l v all lines through 
the centre of an ellipse or hyperbola and all lines parallel 
to the axis of a parabola are diameters. 

Let the student obtain the equation of the diameter of 
each conic by the following method : Transform to polar 
coordinates, with the middle point (x 1 ', y') of any one of 
the parallel chords as origin. If tan~ 1 Z 1 is substituted 
for 6 in this equation of the conic, the resulting values of 
p should be equal in magnitude, but have opposite signs. 
The necessar}^ condition for this will be an equation be- 
tween x r , y\ and l v which will therefore be the equation 
of the diameter. 

80. Conjugate diameters. — If we let l 2 be the slope 
of the diameter which bisects a system of chords of slope 
l v we see that for the ellipse 

7 & J7 V 

and for the hyperbola 

7 & nb 2 



Ch. XL § 80] 



DIAMETERS 



145 



Since in these expressions l x and l 2 are interchangeable, 
it is evident that, if we started out with a system of 
chords of slope 7 2 , the corresponding diameter would have 
l x for its slope. Hence the two diameters ivhich have l x 
and l 2 for their slopes are so related to each other that each 
bisects all chords parallel to the other. Such diameters 
are said to be conjugate to each other. Their equations 

are y = l x x, and y = l 2 x, 

where l x and l 2 are connected by the relations given 
above. In the ellipse, l x and l 2 have opposite signs, and 
the diameters must pass through different quadrants. 
But in the hyperbola, l x and l 2 have the same sign, and 
the diameters must pass through the same quadrant. In 
either case, as l x decreases in numerical value, l 2 increases, 
and as one diameter approaches the major axis, the other 
will approach the minor axis from one side or the other. 
The limiting case is seen to be the two axes. They con- 
form to the definition 
of conjugate diameters, 
since every line paral- 
lel to one is bisected 
by the other. They 
are the onl}' conjugate 
diameters which are 
perpendicular to each 
other. 

If in the ellipse one 

„ * Fig. 79. 

diameter, -r 1 -K v starts 

coincident with the major axis and revolves in the posi- 
tive direction, its conjugate diameter, P 2 K 2 , will start 




146 



ANALYTIC GEOMETRY 



[Ch. XI, § 80 



coincident with the minor axis and also revolve in the 
positive direction, since we have seen that the two must 
pass through different quadrants. 

Let the student show that the angle P x CP 2 s in which 
the minor axis lies, will always be obtuse. 

If, in the hyperbola, P X K X starts coincident with the 
major axis and revolves in the positive direction, its 




conjugate diameter, P 2 iT 2 , will start coincident with the 

minor axis and revolve in the negative direction. For 

the two diameters must remain in the same quadrant. Since 

b 2 b 

the product of the two slopes is — , if l x is less than -, l 2 

must be greater than -, and the two diameters must there- 
a 

fore remain on opposite sides of the asymptote. As l x 

approaches -, L also approaches -, and the asymptote 
a a 

is therefore the limiting position of both diameters. It 



Ch. XI, § 81] DIAMETERS 147 

is evident that only one of tivo conjugate diameters can cut 
the hyperbola. But in speaking of the length of the other 
diameter, we shall mean the distance, P 2 K T between the 
points where it cuts the conjugate hyperbola. 

81. Equation of conjugate diameter. — If one diameter 
is given in any way, as by means of its slope or the coor- 
dinates of its extremity, it is easy to determine the con- 
jugate diameter. For it is only necessary to write the 
equation of a line through the centre whose slope bears 
the required relation to the slope of the given line. If 
the coordinates of P 1 (Fig. 79) are (x v y^), the equa- 
tion of CP 1 is x^y — y x x = 0, and / 2 = — . Then for the 
ellipse, l 

i = _ii= fe i 

2 "% <&/! 

and the equation of the conjugate diameter -P 2 -S" 2 is 

*i? + ^ = 0. [63] 

a 2 b- L J 

The solution of this equation with the equation of the 

ellipse gives for the coordinates of P 9 I ^ — l ), and 

fait bx \ " ^ ^ a ' 
for the coordinates of K -&± l • 

2 V b a J 

Let the student show that in the hyperbola the equa- 
tion of a diameter conjugate to the diameter through 
(as,, y-,) is 

^-^ = 0, [64] 

and that the coordinates of its extremities are 

(Ml, ^ an df-^l, -^l\ 
V b a J V b ' a) 



148 ANALYTIC GEOMETRY [Ch. XI, § 82 

PROBLEMS 

1. Find the equations of a pair of conjugate diameters of 
the hyperbola x 2 — 8 y 2 = 96, one of which bisects the chord 
whose equation is 3 x — 8 y = 10. 

2. Find the equation of the diameter of the parabola 
y' 2 z=Q>x, which bisects all chords parallel to the line x-\-3y = S. 

3. Find the equation of a diameter of the ellipse 

±x 2 + 9y 2 = 36, 
if one end of its conjugate diameter is (fV3, 1). 

4. What is the equation of the chord of the ellipse 
9 x 2 + 36 y 2 = 324, which is bisected by the point (4, 2) ? 

5. Find the equation of a chord of the ellipse 

13^ + 11^ = 113 

through the point (1, 3), which is bisected by the diameter 
2y = 3x. 

6. A diameter of the ellipse 15 y 2 + 4 x 2 = 60 is drawn 
through the point (1, -|). Find the equation of the conjugate 
diameter and its points of intersection with the ellipse. 

7. Find the length of the diameter of the hyperbola 
9 x 2 — 4 y 2 = 36, which is conjugate to the diameter y = 3x. 

8. What is the equation of the chord of the parabola 
y 2 = 6 x, which is bisected by the point (4, 3) ? 

9. A diameter of the hyperbola 4 x 2 — 16 y 2 = 25 passes 
through the point (1, — 2). Find its extremities and the 
extremities of its conjugate diameter. 

10. What is the relation between the slopes of the con- 
jugate diameters of the equilateral hyperbola xy = k? 

82. Theorems concerning diameters. — 1. In the ellipse 
the sum of the squares of any two conjugate semi-diameters 
is equal to the sum of the squares of the semi-axes. 



Ch. XI, § 82] 



DIAMETERS 



149 



In Fig. 81, let CP X = a' and CP 2 = V . Then (by [1]) 



and 



a' 2 = x{ + yf, 

ft/2 _ a W , ft V 
° " 5 2 + a 2 ' 




Fig. 81. 



Adding, a' 2 + 6' 2 = (a 2 + 5 2 )^ + (a 2 + £ 2 )^, 

V 

But since P : is on the ellipse, 

^ + ^ = 1, and a' 2 + i' 2 = « 2 + 5 2 . 

2. 2w Me hyperbola the difference of the squares of any 
two conjugate semi-diameters is equal to the difference of the 
squares of the semi-axes. 

3. The product of the focal distances of any point on a 
central conic is equal to the square of the semi-diameter 
conjugate to the diameter through the point. 



150 ANALYTIC GEOMETRY [Ch. XI, § 82 

Since the focal distances of the point (x v y x ) have been 
shown (Art. 70) to be a + ex 1 and a — ex v it is necessary 
to show that V 2 = a 2 — e 2 x x 2 . From theorem 1 we have 

ft/2 _ g V , ft V 
" J 2 + a 2 * 

But since Pj is a point on the ellipse, 

a 2 y x 2 = a 2 b 2 — b 2 x^, or j\* = a 2 — x^. 

h 2 r 2 

Hence b' 2 = a 2 - x 2 + -^ 



, ~(^^ ? )* ,, ' or (by [89]) ' 



= a 2 — e 2 ^ 2 . 

4. Prove the same theorem for the hyperbola. 

5. The tangents at the ends of a diameter of a central 
conic are parallel to the conjugate diameter. 

In the ellipse the equation of the tangent at P 1 is 

a 2 f b 2 ~ X ' 

This is seen at once to be parallel to the conjugate 

XX 1J II 

diameter -^- + ^Orf =0. In the same way the tangent 

at P 2 can be shown to be parallel to P\K V 

This theorem appears also from the fact that the tan- 
gents are special cases of the system of parallel chords. 

6. The tangent at the end of a diameter of the parabola 
is parallel to the system of chords which the diameter bisects. 

7. The area of the parallelogram formed by tangents at 
the ends of conjugate diameters of a central conic equals the 
area of the rectangle on the principal axes. 



Ch. XI, § 82] 



DIAMETERS 



151 



Let ABED be the parallelogram formed by the tangents 
at the ends of the conjugate diameters P<J£\ anc ^ -^Y^i- 




Fig. 82. 

The sides of the parallelogram are evidently 2 a f and 2 b f . 
From C drop a perpendicular CM on AB. Its length is 
the distance from the origin to the line 

b 2 x x x + a 2 y x y = a 2 b 2 , or (by [IT]) 

car- a2p 



V6 V + a% 2 
But the area of the parallelogram 



= 2 CM x ^5 = 2 (7i!f x P 2 K V 



2 a 2 6 2 



= 4aJ. 



2V&V + ^! 2 
ab 



152 ANALYTIC GEOMETRY [Ch. XI, § 82 

Let the student give the proof for the hyperbola. 
Y 




8. In the hyperbola the parallelogram formed by the 
tangents at the ends of conjugate diameters has its vertices 
on the asymptotes. 

9. In the hyperbola, the line joining the ends of conju- 
gate diameters is parallel to one asymptote and is bisected 
by the other. 

10. Show that the angle between two conjugate diameters 
. i ab 

IS Sin —777- 

a'b' 

11. Conjugate diameters of the rectangular hyperbola are 
equal. 

12. The ellipse has a pair of equal conjugate diameters 
which coincide with the asymptotes of the hyperbola, which 
has the same axes as the ellipse. 



Oh. XL § 82] DIAMETERS 153 

PROBLEMS 

1. Prove that conjugate diameters of an equilateral hyper- 
bola are equally inclined to the asymptotes. 

2. Prove that any two perpendicular diameters of an equi- 
lateral hyperbola are equal. 

3. Prove that the straight lines drawn from any point in 
an equilateral hyperbola to the extremities of any diameter 
are inclined at equal angles to the asymptotes. 

4. Prove that the points on either the major or minor 
auxiliary circle, which correspond to the extremities of a 
pair of conjugate diameters, subtend a right angle at the 
centre of the ellipse. 

5. Prove that chords drawn from any point of a central 
conic to the extremities of a diameter (called supplemental 
chords) are always parallel to a pair of conjugate diameters. 

6. If P, and P 2 are the extremities of a pair of conjugate 
diameters of a central conic, prove that the normals at P 1 and 
P 2 , and the perpendicular from the centre on P X P. 2 meet in a 
point. 

7. If a perpendicular is drawn from the focus of a central 
conic to a diameter, show that it meets the conjugate diameter 
on the corresponding directrix. 

8. Tangents at the extremities of a pair of conjugate diam- 
eters of an ellipse form a parallelogram (Fig. 82). Show that 
the diagonals of this parallelogram are also a pair of conjugate 
diameters. 



CHAPTER XII 

POLES AND POLARS 

83. Harmonic division. — In Art. 14 we have said that 

four points A, B, C, and D on a line form a harmonic 

• f AB AD , , , , 
A B C D ran S e > lf jJQ = - ^* and that 

FlG - 84 - the pairs of points A and C, and 

i? and D are then called conjugate harmonic points. From 
this definition it is easily seen that if we keep A and 
fixed and allow B and D to move, as B approaches (7, D 
will also approach (7, and as B approaches the middle 
point of AC, I) will recede indefinitely. When B coin- 
cides with the middle point of AC, it has no conjugate 
harmonic point. When B moves from the centre toward 
A, B comes in from the left toward A, 

It is desirable for our purposes to express the relation 
between these points in terms of distances from A only. 
From the definition, AB x CD = AD x BC. Substituting 
CD = AD - A C and BC= A C - AB, this becomes 

ABxAD-ABx AC = AD x AC - AD x AB, 

Ari 2AB x AD 
AC= AB + AD ' 

Let the student show that BD = — — =r^r* Connect 

BA + BC 

these results with harmonic progression in algebra by show- 
ing that AC is a harmonic mean between AB and AD. 
154 



Ch. XII, § 84] 



POLES AND POLARS 



155 



84. Polar of a point. — The polar of a point with respect 
to any conic is defined as the locus of points which divide 
harmonically secants through the fixed point. 

The methods of finding this locus are the same for all 
the conies. In problem 31, Chapter VIII, the student 
has been asked to find it for the circle by the aid of polar 




Fig. 85. 



coordinates. The same method might be employed here, 
but it is thought best to use a very similar one, into 
which, however, polar coordinates do not enter. 

We shall find the polar of the point P x with respect to 
the ellipse ^ + ^ y = ^ 

Transform to the point P x as origin by the aid of the 

equations 

X = x + x v 

y = y' + yv 



156 ANALYTIC GEOMETRY [Ch. XII, § 84 

The equation of the ellipse becomes 

b 2 x 2 + a 2 y 2 + 2 J 2 ^ + 2 a 2 */^ + J 2 ^ 2 + a 2 y 2 - a 2 5 2 = 0. 

Let any line, y = Ix, through P x cut the ellipse in the 
points P 2 and P 3 . We wish to find the locus of a point 
P' on this line, so situated that P v P 2 , P\ and P 3 form 
a harmonic range. By the theorem of Art. 14, P v M 2 , 
M, and M z will also form an harmonic range, and hence 

FM= 2P,M 2 xP,M, y=? 2x A 

1 P X M 2 + P X M Z " x 2 + x 3 ' 

If we start the solution of the equation of the line, 
y = lx, with the equation of the ellipse, we have 



(52 + a 2^2) x 2 + 2 (b 2 x x + « 2 ^i) a; + ^i 2 + a 2 ^ 2 - a 2 & 2 = 0, 

Erom which t 
(see Art. 8) 



from which to determine the values of x 2 and x z . Hence 



_ b 2 x^ + a 2 y^ — a 2 b 2 
and * 2 + * 3 = - V +a*/ " 

Hence ^jv +a y-^, 

Since P' lies on the line y = fo, its coordinates satisfy 

the equation, and y' — lx\ or 1 = ^—. Substituting this 

value of I in the equation above, and reducing, we have 

as the equation of the locus, referred to the point P 1 as 

origin, 

b 2 x x x + a 2 y x y + b 2 x^ + a 2 y^ — a 2 b 2 = 0. 



Ch. XII, § 85] POLES AND POLARS 157 

When transformed back to the original origin by the 
aid of the equations x — x' — x x and y = y' — y v this equation 
becomes 

b 2 X\X + a 2 yi V= a 2 b 2 , [65] 

Since this equation is of the first degree, we have 
arrived at the singular result that the locus is a straight 
line. It is called the polar of the point P x with respect 
to the ellipse, while the point P x is called the pole of the 
line. The theory of poles and polars is of great impor- 
tance in some branches of geometry. 

Let the student show that the polar of the point 
(x v y{) with respect to 

(a) the circle, x 2 + y 2 = r 2 , is xix + yiy = r 2 , [66] 

(£>) the hyperbola, 

* b 2 x 2 - a 2 y 2 = a 2 h\ is b 2 x x x - a 2 y x y = a 2 b 2 , [67] 

(<?) the parabola, y 2 = 2 mx, is yiy = mx + mxi, [68] 

(c?) the locus of the general equation, 

Ax 2 + Bxy + Cy 2 + Dx + Ey + F=0, 

is Axxx + — (yix + xiy) + Cyiy 

[69] 

85. Position of the polar. — From what we have said 
about harmonic ranges, it is evident that the polar of every 
point outside the conic cuts the conic, and that the polar 
of every point within the conic does not cut it. The form 
of the equation shows that, tuhen the point is outside 
the conic, the polar coincides with the chord of contact 



158 ANALYTIC GEOMETRY [Ch. XII, § 85 

and, when the point is on the conic, the polar becomes the 
tangent. Again, as P x recedes from the conic, its con- 
jugate harmonic point P' approaches the middle of the 
chord, and its polar, therefore, approaches coincidence 
with a diameter. If the point P 1 is inside a central conic 
and approaches the centre, the polar evidently recedes 
indefinitely. 

PROBLEMS 

1. What is the equation of the polar of 

(a) (1, — 2) with respect to the conic x 2 + 4 y 2 = 16 ? 

(6) (6, — 4) with respect to the conic y 2 = 4 # ? 

(c) (— 3, 2) with respect to the conic 5 x 2 — 8 y 2 = 24 ? 

(d) (0, 0) with respect to the conic 

x 2 +2xy + 3y 2 -±x-l0 = 0? 

2. What is the pole of the line 3x — 2y= 5 with respect 
to the circle x 2 + y 2 = 25 ? • 

Solution. — The polar of the point («i, y{) with respect to the circle 
is X\X + yiy = 25. We wish to find the values of X\ and y x which will 
make this line coincident with the line 3 x — 2 y = 5, or 15 x — 10 y = 25. 
They are evidently xi = 15 and y\ = — 10. 

3. What is the pole of the line 5» + 4t/ = 7 with respect 
to the ellipse x 2 + 2 y 2 = 10 ? 

4. What is the pole of the line x — y — 10 with respect to 
the parabola y 2 = Sx? 

5. What is the pole of the line Ax+By + C=0 with respect 
to the hyperbola iPx 2 — a 2 y 2 = a 2 b 2 ? 

6. Tangents are drawn to the circle y 2 = 10x — x 2 at the 
points where it is cut by the line y = 4 x — 7. What is their 
point of intersection ? 

7. Through the point (a;^ ?/i) a line is drawn parallel to the 
polar of the point with respect to the ellipse IPx 2 + a 2 y 2 = a 2 b 2 . 
Find the coordinates of the pole of this parallel. 



Ch. XII, § 86] 



POLES AND POLARS 



159 



86. Theorems concerning poles and polars. — 1. If a 

set of points lie on a line, their polars all pass through 
the pole of that line ; and conversely, if a set of lines pass 
through a point, their poles lie on the polar of that point. 

Let P 2 be the pole of the line MN, and P x any point 
on MN. We wish to show that RS, the polar of P v 
will pass through P 2 . If the coordinates of P 1 and P 2 
are (x v y x ) and (x 2 , y 2 ), 
the equation of RS is 

IPx^x + a 2 y x y = a 2 b 2 , 

and of MN is 

b 2 x 2 x + <fiy<iy = a 2 b 2 . 

But we know that the 
coordinates of P x must 
satisfy the equation of 

MN, or 

b 2 x 2 x x + a 2 y 2 y 1 = a 2 b 2 . 

Now this is just the condition which must be satisfied, 
if P 2 lies on PS. Hence P 2 lies on MS, and as the point 
P x moves along the line MN, its polar will revolve about 
P 2 , the pole of MN. 

Let the student prove the converse theorem, and also 
both theorems for the hyperbola and parabola. 

It follows from this theorem that tangents at the 
extremities of any chord through P x meet on RS. For 
the pole of every chord through P x lies on RS, and we 
have seen (Art. 85) that tangents at the extremities of 
a chord intersect at the pole of that chord. From this 
property the polar may be defined as the locus of the 




Fig. 86. 



160 



ANALYTIC GEOMETRY 



[Ch. XII, | 



intersection of tangents at the extremities of chords 
through any fixed point. 

This property enables us to construct the polar of any 
point ; for any number of points on the polar may be 




Fig. 87. 



determined by finding the intersections of tangents at 
the extremities of chords through the point. 

2. The polar of any point P x with respect to a central 
conic is parallel to the tangent at the point where the diameter 
through P x cuts the conic. 





—X 



9 Fig. 88. 



Ch. XII, | 



POLES AND POLARS 



161 



Let the coordinates of the point P 2 where CP X cuts the 
hyperbola be (x v y^). Then the equation of the tangent 

2 b 2 x 2 x — a 2 y 2 y = a 2 b\ 

and the equation of the polar of P 1 is 
b 2 x x x — a 2 y x y = a 2 b 2 . 
But since P x and P 2 are on the same line through the 
origin, -i = -?, and these lines are evidently parallel. 
Let the student prove the same theorem for the ellipse. 
3. The polar of any point P x with respect to a parabola is 
parallel to the tangent at the point ivhere a diameter through 
P 1 cuts the parabola. 



Pi,''' 


Y 

/P 2 /Ps 


\ y 


\° / Y 


' 





Fig. 89. 



We may let the coordinates of P 2 be (x 2 ,y^). Then 
the equation of the tangent at P 2 is y x y = mx + mx 2 , and 
the equation of the polar of P x is y x y — mx + mx v These 
equations are seen at once to represent parallel lines. 



162 ANALYTIC GEOMETRY [Ch. XII, § 86 

These two theorems show that the polar of a point on a 
diameter is one of the system of parallel chords bisected 
by that diameter. 

4. If the line joining the centre C of any central conic to 
any point P x cuts the conic in P 2 and the polar of P x in P 3< 
then CP 1 x CP 3 = CP 2 . 

We shall give the proof for the hyperbola, using 
Fig. 88. 

The equation of CP X is y = — x. The coordinates of 

P 2 , where this line cuts the hyperbola are found to be 

Vb 2 x^ — a 2 y^ Vb 2 x x 2 — a 2 y x 2 

and the coordinates of P 3 , where it cuts the polar, 

b 2 x x x — a 2 y x y = a 2 b 2 , 

a 2 b 2 x l , a 2 b 2 y l 

b 2 x x 2 — a 2 y^ b 2 x^ — a 2 y x 2 



Hence CP X = V^ 2 + y 2 , 



2 ~ V b 2 x 2 -a 2 y 2 



_ a 2 b 2 ^x 2 + y 2 
3 ~ b 2 x 2 -a 2 y 2 

From these values we see at once that 

CP 1 x CP Z = CP 2 . 

Let the student prove the same theorem for the ellipse. 
Show that in the parabola (Fig. 89) P 2 bisects the 
line P X P V 



Ch. XII, § 80] POLES AND POLARS 163 

5. The line which joins any point to the centre of a circle 
is perpendicular to the polar of the point with respec$ to the 
circle. 

The proof of this theorem appears at once as soon as the 
equations of the lines are written. This theorem enables 
us to state theorem 4 for the circle as follows : 

6. The radius of a circle is a mean proportional between 
the distance from the centre to any point and the distance 
from the centre to the polar of that point. 

7. The polar of the focus is the directrix in (a) the ellipse, 
(b) the hyperbola, (c) the parabola. 

The proof of this theorem appears at once in each case 
when the coordinates of the focus are substituted in the 
equation of the polar. This theorem is evidently equiva- 
lent to theorems 6 and 7 on tangents. 

8. Any chord through the focus of a conic is perpendicular 
to the line joining its pole with the focus. 

This theorem is equivalent to theorems 11 and 12 on 
tangents and is proved in the same manner. 

9. The line joining the centre of a central conic to any 
point P x cuts the directrix in K. Show that the line KF is 
perpendicular to the polar of P v 

10. Two triangles are so related that the vertices of the 
first are the poles of the sides of the second, with respect 
to a conic. Prove that the vertices of the second are also 
poles of the sides of the first. 

Two such triangles are said to be conjugate to each other. 
If in any triangle the vertices are the poles of the opposite 



164 ANALYTIC GEOMETRY [Ch. XII, § 86 

sides, the triangle and its conjugate coincide, and it is 
called a self-conjugate triangle. 

11. If a line is drawn through a point parallel to the axis 
of a parabola, that portion of it included between the point 
and its polar is bisected by the parabola. 

How does this conform to the definition of harmonic 
division ? 

12. The two lines, ivhich join the focus of a conic to any 
point and to the intersection of the polar of that point with 
the corresponding directrix, are perpendicular to each other. 

13. Write the equation of the polar of a point P 1 on 
a diameter of a central conic. Let P x recede indefinitely 
along the diameter and show that the polar approaches, as its 
limiting position, the diameter conjugate to the given diameter. 
Show that this would be true, if the point receded along 
any line parallel to the given diameter. How must this 
theorem be stated for the parabola ? 

PROBLEMS 

1. Show that the polars of the same point, with respect 
to two conjugate hyperbolas, are parallel. 

2. Show that the four points, in which any line is cut by 
the asymptotes of an hyperbola and by a pair of conjugate 
diameters, form a harmonic range. 

3. What is the polar of the focus of an ellipse, with 
respect to the major auxiliary circle ? 

4. Obtain the equation of the polar of the point P x with 
respect to the rectangular hyperbola xy = k. What are the 
coordinates of the foci and the equations of the directrices 
of this hyperbola? Prove that your results are correct fcy 
showing that the directrix is the polar of the focus. 



Ch. XII, § 86] POLES AND POLARS 165 

5. Show that the polar of one extremity of a diameter of 
an hyperbola, with respect to its conjugate hyperbola, is the 
tangent at the other extremity of the given diameter. 

6. If a perpendicular is let fall from any point P 1 upon 
its polar, prove that the distance of the foot of this per- 
pendicular from the focus is equal to the distance of the 
point P x from the directrix. 

7. An ellipse and an hyperbola have the same transverse 
and conjugate axes. Show that the polar of any point on 
either curve, with respect to the other, is tangent to the 
first curve. 



CHAPTER XIII 
GENERAL EQUATION OF THE SECOND DEGREE 

87. We have seen that the equations of all the conies 
are of the second degree. We shall now prove that an 
equation of the second degree must always represent a 
conic, either in one of the ordinary forms or in one of the 
limiting cases, and show how to reduce any given equation 
to the simplest equation of one of these conic sections. 

88. Two straight lines. — We have seen that there 
are certain equations of the second degree which can be 
factored, and hence represent two straight lines. 

Let us determine what condition must be satisfied by 
the coefficients of the general equation, 

(1) Ax 2 + Bxy + Of + Dx + Ey + F=0, 

when it can be separated into two linear factors. Arrang- 
ing it according to the powers of x and solving, we have 

(2) x = 

- (By +i>) ± V (.£ 2 - 4 AC)y 2 + (2BI) - 4 AE )y + D 2 - 4 AF. 

2A 

If the general equation is to be factored into two 
linear factors, the quantity under the radical must be 
a perfect square. The condition for this is 

(3) (2BD- + AE) 2 - ±(B 2 -±AC)(I) 2 -±AF) = 0, 

or (4) 4 ACE + BI)E - AE 2 - CD 2 - FB 2 = 0. 
166 



Ch. XIII, § 88] EQUATION OF THE SECOND DEGREE 167 

This is, then, the condition which must be satisfied by 

the coefficients of the general equation, when it can be 

separated into two linear factors. The first member is 

called the discriminant of the equation. It is usually 

represented by the letter A. 

Note. — If A = 0, the work will have to be changed somewhat, but the 
same form will always be obtained for the discriminant. 

When this condition is satisfied, the equation can always 
be factored, but it is not necessary that the factors should 
be real. For if B 2 - 4 A is negative, from (3) , D 2 - 4 AF 
must also be negative, and while the expression under the 
radical is a perfect square, its square root will contain 
imaginary coefficients. The equation will in this case 
break up into a pair of factors with imaginary coefficients, 
and we speak of it as representing a pair of imaginary 
lines. There will be, however, one real point on the 
locus ; for the intersection of the two imaginary lines will 
always be a real point. 

If B 2 — 4 AC is positive, the factors represent real and 
intersecting lines. 

If B 2 -±AC=0, 2 BD- 4: AF must also reduce to 
zero, and the quantity under the radical is reduced to 
D 2 — 4 AF. The lines are therefore parallel. They are 
real and distinct if D 2 — 4 AF > 0, 
real and coincident if D 2 — 4 AF = 0, 
imaginary if D 2 — 4 AF < 0. 

PROBLEMS 

1. Obtain the discriminant by the following method: Let 
the two factors be x -f b x y -+- c x and x -f- b 2 y + c 2 . Multiply, and 
equate the coefficients of the product arid those of the general 



168 ANALYTIC GEOMETRY [Ch. XIII, § 89 

equation. This will give five equations from which \, q, b 2 , 
and c 2 can be eliminated and the condition in terms of the 
coefficients obtained. 

2. Show that the following equations represent straight 
lines, and find the factors in each case : 

y- — xy — 5 x + 5 y = 0, 

2 x 2 + 3 xy + y 2 — x — y = 0, 

x 2 + 2xy + y 2 + 2x + 2y + l = 0. 

B 2 -±AC^Q. 

89. Removal of the terms of the first degree. — If the 

discriminant does not vanish, and if the equation does 
represent some conic, it ought to be possible by suitable 
transformations, either by changing the position of the 
origin or by revolving the axes, or both, to reduce it to 
one of the well-known forms. 

Let us transform to a new origin (# , y ) and find the 
values of x and y^ if any, which will simplify the equa- 
tion. The general equation becomes 

(5) Ax 2 + Bxy + Of + D'x + E'y + F = 0, 
where (6) D' = 2 Ax + By + i>, 

(7) E< = Bx + 2Cy + E, 
and (8) F = Ax* + Bx y + Cy 2 + JDx 4- .% + ^. 

It appears then that we can choose x and y so that 
any two of the last three terms shall vanish. Let them 
be chosen so that B ! and E' shall be zero, or so as to 
satisfy the two equations, 

(9) 2Ax + By + B = 0, 
and (10) Bx + 2Cy + E=0. 



Ch. XIII, § 90] EQUATION OF THE SECOND DEGREE 169 

™ 2CD-BJE. , 2AU-BB 

Phen ' **= &-1AC ^ y ° = #-*AC ' 

The general equation is reduced by this transformation 

(11) Ax 2 + Bxy + C y f + F' = 0, 

in which the value of i^' is found by substituting x and 
«/ for x and ?/ in equation (1). 

But, if B 2 — 4 A 0= 0, no values can be found for x 
and £/ , and this transformation is therefore impossible. 
Let this case be set aside for the present, and only those 
cases be considered where B 1 — 4 AC ^= 0, and where this 
transformation is therefore possible. 

We have reduced the equation to a form which shows 
that the curve is symmetrical with respect to the origin, 
for any line, y = Ix, through the origin meets it at two 
points equally distant from the origin. But the term in 
xy must be removed before it is symmetrical with respect 
to the axes. 

90. Removal of the term in xy. — Let the axes be 
revolved through any angle 6 by substituting 

x = x f cos — y' sin 0, 

y = x' sin 6 + y 1 cos 6. 

The equation now becomes 

(12) A'x* + B'xy + C'y* + J" = 0, 

where (13) A' = A cos 2 6 + B sin 6 cos 6 + C sin 2 0, 

(14) £' = (<?- A)sin2 + £cos2 0, 

(15) C = A sin 2 - 5 sin (9 cos 6 + Ccos 2 0. 



170 ANALYTIC GEOMETRY [Ch. XIII, § 91 

Evidently 6 can be so chosen as to make any one of 
the three coefficients zero. But it is the term in xy 
which is not wanted. We shall then choose so that 
£'=0. 

B 



Hence (16) tan 20 = 



■0 



There will always be two values of #, one acute and 
the other obtuse, which will satisfy this equation. But, 
for the sake of uniformity, we shall always choose the 
acute value. 

The equation will be reduced by this transformation to 

(17) A'x 2 + C'yi + F = 0. 

91. Determination of the coefficients A\ C ', and F' . — We 
have shown how to determine F ! ; and since tan 2 6 is 
known, the values of A! and C may be found, and the 
result fully determined. But much of the labor involved 
may, in practice, be avoided by the following method : 

Adding equations (13) and (15), we have 

(18) A' + CP = A + 0. 

Subtracting the same equations, we have 

(19) A'-C =(A- C)cos20 + Bsm26-. 

Squaring (19) and (14), and adding, we have 

(20) (A! -C f y + B ,2 = (A-C) 2 + B 2 . 

Squaring (18) and subtracting from (20), we have 

(21) B^-4,A'C' = B^-4:AC. 

These results hold good for all transformations from 
one system of rectangular axes to another. 



Cn. XIII, § 91] EQUATION OF THE SECOND DEGREE 171 

If the general equation has been reduced to equation 
(17), B' = 0, and (21) reduces to 

(22) ±A f C' = 4 AC -B 2 . 

From the two equations (18) and (22), A' and C can 
be found. But there will be two values of each, corre- 
sponding to the two possible values of 0, and it will be 
necessary to be able to choose the proper values. 

We have let (C- A) sin 2 6 + £cos 2 6 = 0. 

Multiplying this equation by (J. — (7) and equation 
(19) by B and subtracting, we have 

(23) (B 2 + (A - 6 7 ) 2 )sin 2 = B(A' - <?')• 

If now the acute value of be chosen, the first member 
will always be positive, and the factors of the second 
member, B and A' — C , must have the same sign. It will 
be easy then to choose the proper values for A' and C . 

The determination of F' may also be considerably sim- 
plified. Multiply equation (6) by x and (7) by y Q and 
add. The sum is 

2^V+ 2 ^o + 2 W + ^o + %o = °- 
Combining this with (8), we have 

i" =f*o+ !*<>+* 

Substituting the values of x and «/ , 

r24 . F ,_ CD 2 + AE 2 -BDE+B 2 F-±ACF ^ -A 
^ ; B*-±AC B 2 ~4AC 



172 ANALYTIC GEOMETRY [Ch. XIII, § 92 

92. Nature of the locus. — The general equation has 
now been reduced by transformation of coordinates to 
the form 

(17) A'x* + c y 2 + F' = 0. 

Neither of the coefficients A' or C can be zero, for 
they must satisfy equation (22), and we are only con- 
sidering the case where B*-4:AC=f=0. If F' =£ 0, (17) 
can be written in the form 

71 " 1 " 



— F' -F'~ 

~A r "C^ 

The nature of the curve evidently depends on the rela- 
tive signs of A 1 , C, and F' . If A! and C have the same 
sign and F' has the opposite sign, equation (17) will 
represent a real ellipse; for it can be written in the 

form 2 2 

x ± + f = \ 
a 2 b 2 

If A', C", and F' all have the same sign, equation (17) 
can be written in the form 

a 2 ^b 2 

This equation has no real locus, but is said to represent 
an imaginary ellipse. 

Again, if A 1 and C have opposite signs, the equations 
will take one of the two forms 

<L TM. — ^1 Ui ~2 Z.2 X ' 



Ch. XIII, § 92] EQUATION OF THE SECOND DEGREE 173 

according as F' has the same sign as C or as A'. These 
are both real hyperbolas. 

From equation (22), 4 A' C = 4 AC- B 2 , we see that 
A' and C have the same or opposite signs according as 
B 2 — 4 A C is negative or positive. If then F' is not zero, 
or what is the same thing, if the discriminant does not 
vanish, and if B 2 — 4: AC =£ 0, the general equation has 
been shown to represent 

an ellipse, real or imaginary, when B 2 — 4 AC < 0, 

an hyperbola, always real, when B 2 — 4 AC> 0. 

If F' = 0, equation (17) reduces to 

(25) A'x 2 + C'y 2 = 0. 

When A' and C have the same sign, the equation may 
be looked upon as representing a pair of imaginary lines, 
since the equation can be separated into a pair of linear 
factors with imaginary coefficients. These lines have a 
real point of intersection, the origin. Or it may be looked 
upon as the equation of an ellipse in which the axes have 
become zero. From this point of view, it is spoken of as 
representing a null ellipse. 

When A' and C r have opposite signs, the equation can 
always be separated into two real factors, representing 
two real and intersecting lines. 

These results will be seen to agree with those obtained 
in Art, 88. 

PROBLEMS 

1. Determine the character of the locus of the following 
equation, reduce it to its simplest form, and plot : 

5 x 2 + 2xy + 5y 2 - 12 a - 12 y = 0. 



174 ANALYTIC GEOMETRY [Ch. XIII, § 92 

Substituting these values for the coefficients in (4), we 
obtain A = -1152. Also B 2 - 4 AC = -96. The locus is 
therefore an ellipse, real or imaginary. 

The simplest method for determining the coordinates of the 
centre is to write the equations (9) and (10) and solve for 
x and y . 

The first of these may be obtained by multiplying the coeffi- 
cient of every term which contains x by the exponent of x, 
decreasing that exponent by unity, and leaving out all terms 
which do not contain x. The second may be formed in a 
similar way, using y. In this case they are 

10 x + 2y - 12 = 0, and 2 x + 10y -12 = 0. 

From these the coordinates of the centre are found to be (1, 1). 
From equation (24), F' =— 12. The equation, referred to 
the point (1, 1) as origin, is then 

5x 2 + 2xy + 5y 2 -12 = 0. 

Next revolve the axes through an angle 0, such that 

tan 2* = -^=^. 

We have decided to use the acute value of 6, which is here -• 

4 

To determine A' and C, we use the equations (18) and (22) 

or ' A' + C = A + C = 10, 

and <LA'C' = 4;AC-B 2 = 96. 

Solving, we have A' = 6 or 4, and C = 4 or 6. But since 
we chose the acute value of 6, we must choose A' and C so that 
A' — C has the same sign as B. This is positive. Hence the 
final form of the equation is 

6.r 2 +4?/ 2 = l2. 



Ch. XIII, § 93] EQUATION OF THE SECOND DEGREE 175 




But this is the equation of the curve referred to axes 
with origin at the point 
(1, 1), and making an angle 

of - with the original axes. 

4 

Constructing these axes 
and plotting the equation 

6ar 9 + 4?/ 2 = 12 
with respect to them, we 
have the locus of the origi- 
nal equation, referred to the 
original axes. 

2. Determine the char- FlG - 90 - 

acter of the loci of the following equations, reduce them to 
their simplest forms, and plot : 

(a) 2x 2 + 2y' 2 - Ax-Ay + 1 = 0, 

(6) x 2 + tf + 2x + 2 = 0, 

(c) 4 xy -235 + 2 = 0, 

(rf) y 2 — 5 .17/ + 6 x 2 — 14 x + 5 y -f 4 = 0, 

B 2 - 4 AC = 0. 

93. Removal of the term in xy. — We have seen that, if 
W — 4 A C = 0, it is not possible to transform to a new 
origin such that the terms in x and y shall disappear. In 
this case we shall first revolve the axes through an angle 0. 

Proceeding as in Art. 90, we obtain the equation 

A'x 2 + B'xy 4- Cif + D'x + E'y + F=0, 
where (13) A' = A cos 2 + B sin cos + Csin 2 0, 

(14) B' = (C-A) sin20 + Bcos20, 

(15) C = A sin 2 - B sin cos + Ccos 2 0, 

(26) B' =Dcos0 + .#sin<9, 

(27) E' =- D$in0 +Eeos0. 



176 ANALYTIC GEOMETRY [Ch. XIII, § 93 

The values of A r , B', and C are the same as those used 
in Art. 91. The results there obtained will therefore 
apply here. These were, 

(18) A 1 + C = A + C, 

and (21) B' 2 -4:A'C' = B 2 -4:AO. 

TO 

Let 6 be so chosen that B' = 0, or tan 2 6 = — Then 

A — O 

since B 2 — kAC — 0, it appears from (21) that either A! 
or C must reduce to zero at the same time. It can easily 
be shown that one of the two values of will give A! = 0, 
and the other, C = 0. Let that value be chosen which 
will make 

A' = A cos 2 + B sin 6 cos 6 + (7sin 2 = 0. 

Solving, we have A + B tan + # tan 2 = 0, 
(28)ta„tf=-^ = -?A 

The general equation will be reduced by this transforma- 
tion to 

(29) C'y* + D'x + E'y + F=^ 

where (30) C'=A+C, 

(31)2>'= BD-2AE 

and (32) ..#' = 



±VB* + ±A 2 

EB + 2 AD 

±V^ + 4A 2 



It appears then that C" cannot vanish, since A and 
(7 have the same sign ; that B' or E may vanish, but 
since BD — 2 J.i7 is the value of the discriminant when 
B 2 — 4 AC= 0, i)' cannot be zero unless A = 0. 



Ch. XIII, § 95] EQUATION OF THE SECOND DEGREE 177 

94. Removal of the term in /. — Transform equation 
(29) to a new origin (z , y^). It becomes 

(33) C'y 2 + D'x + E"y + F f = 0, 

where (34) E" = 2 C'y + E\ 

and (35) F' = C'y* + 2>'# + E% + .F. 

We can then, in general, choose such values for x and y Q 
that E" = F' = 0. Solving the two equations 

and C'y* + 2>% + -# '^o + ^ = °> 

, ^' , E'*-4:C f F 
we have y = - — , and s = 4 ^ j>r 

If D' ^=. 0, these values are always finite, and the trans- 
formation is possible. The equation will be reduced by it to 

(36) C'y 2 + D'x = 0. 

If D' = 0, no value can be obtained for x which will 
make F' = 0. But if we transform to the point (0, y ), 
the equation will be reduced to 

(37) C'y 2 + F' = 0. 

95. Nature of the locus. — When B 2 -±AC=0, the 

general equation has been reduced by transformation of 
coordinates to one of the two forms 

(36) C'y 2 + D'x = 0, or y 2 = - ^x, 

(37) CV + *' = 0,OTj» = - : jj[. 



178 ANALYTIC GEOMETEY [Ch. XIII, § 96 

The first of these equations always represents a real 
parabola. The second is obtained only when A = 0, and 
represents, as we should expect, a pair of lines. In this 
case the lines are evidently parallel and 

real and distinct, if C and F ! have opposite signs, 

real and coincident, if F' = 0, 

imaginary, if C'and F' have the same sign. 

It has been shown that 

(30) C'=A+C, 

and (31) D' = ^ >- 2 ^ _ 

±VB 2 + 4A 2 

and when A = 0, it can be shown that 

4 AF - Z> 2 



(38) F' . 



±A 



From these values the reduced form of the equation can 
be determined. But in any numerical problem the method 
of the following section will be found to be simpler. 

PROBLEM 

1. Show that the above conditions which determine the 
nature of the parallel lines are the same as those given at the 
end of Art. 88. 

96. Second method of reducing the general equation to a 
simple form, when B 2 -±AC = 0. — When B 1 - 4 AC = 0, 
the terms of the second degree in the general equation 



Ch. XIII, §96] EQUATION OF THE SECOND DEGREE 179 

form a perfect square, and the equation can be written in 

the form 

(ax + cy) 2 + Dx + Ey + F= 0, 

where a = ^/A and c = V(7. 

Introduce arbitrarily the quantity k inside the paren- 
thesis, and subtract from the rest of the equation whatever 
has been added by this introduction. It becomes 

(ax + cy + k) 2 + (D - 2 ah)x + (E - 2 ck)y + F - k 2 = 0. 

Now choose such a value for k that the two lines repre- 
sented by the equations 

ax -\- cy + k = 

and (B- 2 ak~)x + (E- 2 ck)y + F-k 2 = 

shall be perpendicular to each other. Let I be such a 
value of k. The equation will then take the form 

(ax + cy + 2 = D'x + E'y + F' . 

Divide both members of this equation by a 2 + c 2 , 
and both divide and multiply the second member by 
~VD' 2 + E' 2 , and write the result in the form 

f ax + cy + l \ 2 = Vi)' 2 + E' 2 f D'x + E'y + F' \ m 
\ ^/W+J 2 J a^ + c 2 \ VD' 2 + E' 2 J 

But y is the distance of the point (x, y*) from 

Va 2 + c 2 
the line ax + cy -h 1 = 0. Represent this by y' '. 



180 ANALYTIC GEOMETRY [Ch. XIII, §96 

Again, ^ — is the distance of the point 

VD' 2 + E' 2 F 

(x, y) from the line D'x + E'y -f F' = 0. Represent this 
by x f . Then the equation reduces to 



where x' and y' represent the perpendicular distances of 
any point on the locus from the two perpendicular lines 

D'x + E f y + F , = 

and ax -f- cy + I = 0. 

It is therefore the equation of the curve referred to 
these lines as Y and X-axes respectively. The positive 
direction of the X-axis can be fixed by finding the inter- 
cepts of the curve on the original axes, and determining 
by inspection which way the parabola is turned. 

PROBLEMS 

1. Plot the locus of the equation 

x 2 - 2 xy + if - 8 x + 16 = 0. 

Following the method described above, write the equation in 
the form 

(x-y + Jcf -(8 + 2 k)x + 2 ky + 16 - k 2 = 0. 

If the two lines represented by the equations 

x — y + & = 

and -(8 + 2 k)x + 2 ley + 16 - k 2 = 



Ch. XIII, § 96] EQUATION OF THE SECOND DEGREE 



181 



are perpendicular to each other, k = — 2. Substituting this 
value in the equation and transposing, it becomes 

( x -y-2y = 4(x + y-3). 

Dividing both members by a 2 + c 2 , and both dividing and 
multiplying the second member by Vi>' 2 + E' 2 , it becomes 



V2 



:2V2^ 



V2 



or y< 



2-V2q 



where y' is the perpendicular distance of any point (x, y) of 

the locus from the line x — y — 2 = 0, and where x' is the 

distance from x-\-y— 3 = 0. 

It is therefore the equation 

of the locus referred to 

these lines as X and Y- 

axes. 

Construct the two lines. 
From the original equation 
we see that the curve 
touches the X-axis at 
the point (4, 0), and does 
not cut the F-axis. It 
is then easily seen which FlG - 91 - 

is the positive direction of the axis O'X', and the curve can be 
plotted as in Fig. 91. 

2. Plot by this method the locus of the following equations: 
(a) x 2 -2xy + y 2 -6x-6y + 9 = 0, 

(6) x 2 + 6 xy + 9 y 2 + x - 6 y -9 = 0, 

(c) 2 x 2 + 8 f + 8 xy + x + y + 3 = 0, 

(d) y 2 - 2 x - 8 y + 10 = 0, 
(c) ±x 2 + 4:xy + 2j 2 + 6 = 0, 




182 . ANALYTIC GEOMETRY [Ch. XIII, § 97 

97. Summary. — It has been shown in this chapter that 
the general equation of the second degree represents, 

and when B 2 — 4 A C < 0, an ellipse (real or 

i a n imaginary), 

when A * 0, and when w _ 4 A Q = Q? ft parabola? 

and when J9 2 — 4 ^1 C > 0, an hyperbola ; 

and when B 2 — 4 AC < 0, a null ellipse (two 
imaginary lines), 

and when B 2 — 4 J. C = 0, two parallel lines 
wnen a = , (real, coincident, or imaginary), 

and when B 2 — 4 A C > 0, two real intersect- 
ing lines. 

All of these forms may be obtained as plane sections of 
a right circular cone, and are all included under the term 
"conies." An equation of the second degree must therefore 
represent some conic either in its regular or degenerate form. 

PROBLEMS 

Determine the nature of the locus of each of the following 
equations : 

1. 3x 2 -2xy + y 2 + 2x + 2y + 5 = 0. 

2. x 2 + xy + y 2 + 2 a + 3 y -3 = 0. 

3. 2a 2 -5a?/-3?/ 2 + 9a-13 2/ + 10 = 0. 

4. 4a 2 + 2a?/-?/ 2 + 6a + 2?/ + 3 = 0. 

5. 9 x 2 - 12 xy + 4y 2 - 24.x + 16 y- 9 = 0. 

6. 9x 2 -6xy + y 2 + 4 a + 3 y + 16 = 0. 

7. 25 x 2 + 40 xy + 16 y 2 + 70 x + 56 y + 49 = 0. 

8. 13 x 2 + 14 ay + 5 y 2 + 14 a + 10 y + 5 = 0. 

9. 4a* + 9y 2 - Sx + 54?/ + 85 = 0. 

10. 3 a 2 + 10 ay + 7 y 2 + 4 a + 2y + 1 = 0. 



Ch. XIII, §99] EQUATION OF THE SECOND DEGREE 183 

98. General equation in oblique coordinates. — If the 

general equation is referred to axes which are oblique, 
we can first transform to rectangular axes with the same 
origin. The resulting equation will be in the form 

A'x + B'xy + C'y + D'x + E'y + F' = 0. 

This can then be treated by the methods of this chapter. 
It must, therefore, represent a conic. 

99. Conic through five points. — The general equation 
of the second degree contains six constants, but only five 
of these are independent, since any one we please may be 
reduced to unity by division. Five conditions are therefore 
sufficient to determine the conic. For example, it can be 
made to pass through five points, and in general no more 
than five. For, if the coordinates of the five points are 
substituted in turn in the general equation, there will be 
five equations from which, in general, we can determine 
five coefficients in terms of the sixth, which will divide 
out after substitution. If a sixth point were given, there 
would be six simultaneous equations in five variables, 
which is not possible unless some of the equations are not 
independent. This will only happen when the sixth point 
lies on the conic through the other five. 

If three of the points lie on a line, the conic evidently 
breaks up into this line and another line through the other 
two. If four points lie on a line, the solution is indeter- 
minate ; for this line and any other through the fifth 
point will be a conic through the five given points. 

Other conditions may be given, as in the case of the 
circle, where A = C and B = 0. These two conditions 



184 ANALYTIC GEOMETRY [Ch. XIII, §99 

restrict the number of points through which a circle can 
be passed to three. Similarly, a parabola can be passed 
through only four points, since the condition B 2 — 4 A (7=0 
must be satisfied. But here, since the condition is a quad- 
ratic, there may be two parabolas which pass through the 
four points ; or imaginary solutions may be obtained, and 
four points may therefore be chosen through which no real 
parabola can be drawn. 

PROBLEMS 

1. Find the equation of a conic through the points 

(a) (2, 3), (0, - 3), (2, 0), (5, 5), (- 5, - 5). 
(6) (5, 3), (4, 4), (2, 6), (7, 1), (0, 0). 
(c) (2, 4), (4, 3), (6, 2), (0, - 1), (1, 0). 

2. Find the equation of a parabola through the points 

(a) (0, 0), (8, 8), (4, 2), (- 4, 2). 

(b) (0,0), (1,0), (-1,1), (-1,-1). 
(e) (4,3), (0,-4), (6,1), (-6,2). 
(cf) (12,-6), (3,0), (0,2), (-3,4). 

3. Determine the nature of the conies obtained in problems 
1 and 2. 



CHAPTER XIV 
PROBLEMS IN LOCI 

1. Find the locus of the vertex of a right angle whose 

sides are tangent to the ellipse — + %■ — 1. 

a 2, b 

The equations of any two perpendicular tangents P'K 
and P'L may be written in the form 



> = l 1 x + Vl 1 2 a 2 + b 2 , 



and 



y = I 2 x + v7 2 % 2 + b 2 , 



where ? x / 2 = 



1. If P' is their point of intersection, its 
coordinates (V, y) must 
satisfy both equations. 

Substituting these co- 
ordinates, and replacing 
l 2 by , we have 



r y' = l x x l + ^/l 2 a 2 + b 2 , 



i i 



-f*+V£+ 



h 



I 2 




Fig. 92. 



two equations in x\ y\ 
and the variable param- 
eter l v By eliminating l v we shall obtain a single equa- 
tion in x f and y f . Clearing of fractions, transposing, and 
squaring, 

185 



186 ANALYTIC GEOMETRY [Ch. XIV 

y' 2 - 2 l lX 'y' + h 2 x' 2 = l^a 2 + 6 2 
7^' 2 + 2 Z^y 4- x' 2 = a 2 + ^ 2 
Adding, (l + ^z/^ + Cl + ^^^Cl + ^^^ + Cl-h^ 2 )^. 
Dividing by (1 + Z x 2 ), y' 2 + a/" = a 2 + 6 2 , 
or .t 2 + y 2 = a 2 + b 2 . 

The locus is the director circle, a circle having the same 
centre and Va 2 + b 2 as radius. 

2. Find the locus of the intersection of perpendicular 
tangents to a parabola. 

3. Find the locus of the intersection of tangents to the 
ellipse if the product of their slopes is constant. 

As in problem 1, the equations connecting x\ y\ I v and 
l 2 are 

(1) y ' = l lX ' + ^/l 2 a 2 + b 2 , 

(2) y f = l 2 x'+Vl 2 2 a 2 +b 2 , 

(3) y a = *. 

But the method of elimination used in that problem 
will not apply here. Transpose and square (1) and (2), 

y' 2 - 2 l^x'y' + lfx n = l 2 a 2 + 5 2 , 

y' 2 - 2 l 2 x'y' + l 2 x' 2 = l 2 a 2 + b 2 . 

Write these as affected quadratic equations in 7 2 and Z 2 , 

(4) ( a 2 _ ^2) ^2 + 2 a;y l x + 6 2 - / 2 = 0, 

(5) (a 2 - z' 2 ) Z 2 2 + 2 a/y'Jj, + 6 2 - y' 2 = 
If now we write the equation 

(6) O 2 - z' 2 ) 2 2 + 2 x'y'z + b 2 -y' 2 =0 



Cn. XIV] PROBLEMS IN LOCI 187 

(an affected quadratic in z), it appears from (4) and (0) 

that l x and l 2 are the two roots of (6), and hence that 

b 2 — y' 2 b 2 — y' 2 

hh = ,o - But hh = &. Hence — — ^— = k is the 

a 1 — x J a 1 — x l 

equation of the desired locus. Dropping primes and 
reducing, we have kx 2 — y 2 = ka 2 — b 2 . 

If k = — 1, it becomes # 2 + y 2 = a 2 + 6 2 , as in problem 1. 

4. Find the locus of the intersection of tangents to the 
parabola if the product of their slopes is constant. 

5. Find the locus of the feet of perpendiculars from 
a focus on tangents in the (a) ellipse, (5) hyperbola, 
(V) parabola. 

6. Find the locus of the intersection of tangents at the 
ends of conjugate diameters of an ellipse. 

Note. — Solve this as a special case of problem 3. 

7. Find the locus of the intersection of tangents at 
the ends of conjugate diameters of an hyperbola. 

8. Radii vectores are drawn at right angles from the 
centre of an ellipse. Find the locus of the intersection of 
tangents at their extremities. 

9. Find the locus of the middle point of chords joining 
the ends of conjugate diameters of an ellipse. 

Let (.V, y) be the middle point of any such chord. If 

(. X \V\) are tne coordinates of P r ( ~ ?^\ — ) will be the 
coordinates of P 2 . 

Xl b Vl+ n 

Then x' = ^ , and y' = - 



2 ' * 2 

or (1) 2 bx' = bx x — ay v 

and (2) 2 ay' = ay x + bx v 



188 ANALYTIC GEOMETRY- [Ch. XIV 

Since (x^^) lies on the ellipse, 

(3) b 2 x 2 +a 2 y 2 = a 2 b 2 . 

From these three equations we can obtain a single 

equation in x' and y' by 
eliminating x 1 and y v 
From (1) and (2), 




bx' 



■ay . 



. ay 



bx' 



Fig. 93. 



Substituting these val- 
ues in (3), it reduces to 



10. Find the locus of the vertex of a triangle whose 
base is a line joining the foci and whose sides are parallel 
to two conjugate diameters. 

11. Find the locus of the middle point of chords 
drawn through a fixed point in the (a) ellipse, (5) 
parabola. 

12. Tangents are drawn to the parabola y 2 = 2mx. 
Find the locus of their pole with respect to the circle 
x 2 + y 2 = r 2 . 

13. The two circles x 2 + y 2 — a 2 and x 2 + y 2 — ax = 
are tangent internally. Find the locus of the centres 
of circles which are tangent to both the given circles. 



Ch. XIV] 



PROBLEMS IN LOCI 



189 



Let the two circles be drawn, and let (V, 2/') be the 
centre of any circle which is tangent to both circles. 
Then the lines O'P' must 
pass through B, the point 
of contact of the two 
circles, and OP' must 
pass through C. Hence, 
p< 0=00 -OP' 

= r - OP', 
and P'B = P'0'-BO' 

=P'0'-L 



But P' C and P' B are 

radii of the same circle. 





Y 






10 0' 




\ / / 


W"' / 




M fo 


/ 




c^^i 







Fig. 94. 



Hence, 



OP' =P'0'-~, 
2 



y-V^TP 



=V(* ; 



+ y"< 



Squaring and reducing, the equation of the locus re- 
duces to 8 x 2 + 9 y 2 — 4 rx — 4 r 2 = 0. What curve is this 
and how is it situated? 

14. Find the locus of the centres of all circles which 
pass through the point (0, 3) and are tangent internally 
to x 2 + y 2 = 25. 

15. Find the locus of the centres of circles which are 
tangent to a given circle and pass through a fixed point 
outside of that circle. 

16. Lines are drawn from the point (1, 1) to the 
hyperbola x 2 — y 2 = 1. Find the locus of the points which 
divide these lines in the ratio of 2 to 1. 



190 



ANALYTIC GEOMETRY 



[Ch. XIV 



17. Lines are drawn from the centre of the circle 
x 2 + y 2 = r 2 , cutting the circle in A and the line, x = a, 
in B. Find the locus of P, if 0, A, B, and P form a 
harmonic range. Show that the result will represent an 
ellipse, hyperbola, or parabola, according as 4 r 2 < a 2 , 
4 r 2 > a 2 , 4 r 2 = a 2 . 

18. Find the locus of the vertex of a triangle if the 
length of the base is <?, and the product of the tangents 
of the base angles is k. 

Let P' be any position of the vertex of the triangle, 
and 00 the base. We know that 

tan OP' . tun P' CO = k. 



But tan COP = - 



and tanP'C0 = — *— r 

c — x 

Hence the condition which 
must be satisfied is 

,/2 




Fig. 95. 



jr 



-=*. 



a/(tf — x l ) 

Dropping primes and reducing, we have 
kx 2 + y 2 — hex = 0. 

This will be an ellipse or hyperbola, according as k is 
positive or negative. In either case the coordinates of 
the centre will be { -, ), and the semi-axes will be - and 
cVk 

T~' 

19. Find the locus of the vertex of a triangle if the 
length of the base is c, and the product of the tangents 



Ch. XIV] PROBLEMS IN LOCI 191 

of the half base angles is k. Show that the locus is an 
ellipse with the extremities of the base as foci. 

Note. — Express the tangents of half the base angles in terms of the 



three sides by the aid of the formula tan \A =\— — — — ■ 

> s(s - a) 

20. Find the locus of the vertex of a triangle if the 
length of the base is c, and one of the base angles is twice 
the other. 

21. Find the locus of the intersection of tangents to the 
(a) parabola, (5) ellipse, (c) hyperbola, which include an 
angle 6. 

Show that if 6 — 90°, the results reduce to those ob- 
tained in problems 1 and 2. 

22. Find the locus of the centre of a circle which 
passes through a fixed point and touches a given line. 

23. A straight line, whose length is c, slides between 
two perpendicular lines. Find the locus of the intersec- 
tion of the medians of the triangles formed. 

24. If a straight line passes through a fixed point, 
find the locus of the middle point of that portion of it 
intercepted between two perpendicular lines. 

25. Tangents are drawn to a circle from a variable 
point on a given fixed line. Prove that the locus of the 
middle point of the chord of contact is another circle. 

26. Find the locus of the intersection of a tangent to 
the circle x 2 + y 2 = a 2 , and a perpendicular on the tangent 
from the point (a, 0). 

27. Find the locus of the poles, with respect to the 
parabola y 2 =2mx, of tangents to the parabola y 2 = — 2mx. 



192 



ANALYTIC GEOMETRY 



[Ch. XIV 



28. Find the locus of the middle points of chords of 
an ellipse whose poles lie on the auxiliary circle. 

29. Find the locus of the intersection of two perpen- 
dicular lines which are tangent respectively to two con- 
focal ellipses. 

Let the equation of the 
two ellipses be 

W ^ + fs = 1 < 

a 1 b 2 




( 2 > 4+f5 =i - 



Since they are confocal, 
the value of c will be the 
same in both. Hence 

(3) a 2 -b 2 = a 2 -b 2 . 

The equations of the tangents to (1) and (2) are 



Fig. 96. 



y = lx + VPa 2 + b 2 , 



■A^ + V- 

Let (V, ?/) be their point of intersection. Then 



»--!+- 



y = fo'+v7 2 a 2 + & 2 , 



-?+v 



- + b 2 . 



l M 2 

The elimination of I will give a single equation in x' 
and y'. Transpose and square. 

y' 2 - 2 Ix'y' + l 2 x' 2 = l 2 a 2 + b 2 
Py f2 + 2lx'y r + x' 2 = a 2 + b 2 l 2 

Adding, (1 + l 2 )y' 2 + (1 + Z 2 >' 2 = «i 2 + <^ 2 + J2 + ^i 2 ^ 2 



Ch. XIV] PROBLEMS IN LOCI 193 

But from (3), a 2 + b 2 = a 2 + b 2 . 

Substituting and factoring, we have 

(1 + l 2 )y' 2 + (1 + l 2 )x' 2 = a 2 (l + I 2 ) + ^ 2 (1 + Z 2 ), 
y' 2 + x' 2 = a 2 + b 2 , ' 

or, dropping the primes, we have for the equation of the 

locus 

x 2 + y 2 = a 2 + b 2 . 

30. Find the locus of the intersection of two perpen- 
dicular lines which are tangent respectively to two con- 
focal parabolas. 

Note. — Write the equation of the parabola referred to the focus as 
origin, y 2 = 2 mx + m 2 , and obtain the equation of the tangent to it in 

terms of the slope, y = Ix + w (* + ?2 \ 

31. Find the locus of the points of contact of tangents 
drawn from a fixed point on the principal axis to a set 
of confocal ellipses. 

32. Find the locus of the middle points of -chords in a 
circle, which are tangent to an internal concentric ellipse. 

Let (V, ?/') be the middle point of the chord, and 
(x v y{) and (x v y 2 ) its extremities. Then the equation 
of the chord will be 

The condition which makes this line tangent to the 
ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 is 

Since (x x y^) and (z^) are on the circle, 



194 ANALYTIC GEOMETRY [Ch. XIV 

(2) x*+ij* = r\ and (3) x* + yg = r 2 . 

Also, (4) x' = *jJ^2, and (5) y' = l^ti??. 

From these five equations we can eliminate x v y v a? 2 , 
and y v and obtain a single equation in x' and y\ which 
will be the equation of the locus. 

33. Given two concentric ellipses, one within the other, 
on the same axes. Find the locus of the pole of tangents 
to the inner with respect to the outer. 

34. Find the locus of the middle points of a set of 
parallel chords intercepted between an hyperbola and its 
conjugate. 

35. Normals are drawn to an ellipse and the circum- 
scribing circle at corresponding points. Find the locus 
of their point of intersection. 

36. A perpendicular is drawn from a focus of an ellipse 
to any diameter. Find the locus of its intersection with 
the conjugate diameter. 

37. Find the locus of the middle point of all chords of 
an ellipse of the same length 2 c. 

Note. — Find the polar equation of the ellipse referred to the point 
(V, y') as origin. Then express the conditions that the two values of p 
are each equal numerically to c, but opposite in sign. Eliminate 0. 

38. Find the locus of the intersection of the ordinate 
of any point of an ellipse, produced, with the perpendicu- 
lar from the centre to the tangent at that point. 



PART II 

ANALYTIC GEOMETRY OF SPACE 
CHAPTER I 

COORDINATE SYSTEMS. THE POINT 

1. In the following chapters on Analytic Geometry of 
Space, a knowledge of the methods and results of Solid 
Geometry and of Plane Analytic Geometry is presumed. 
Many of the methods and formulas to be given for three 
dimensions are closely analogous to methods and formulas 
in two dimensions, with which the student is already 
familiar ; and in all such cases the discussion will be 
condensed into as brief a form as possible. 

For convenience of reference, the following theorems 
and definitions from solid geometry are cited : 

If a straight line is perpendicular to a plane, it is per- 
pendicular to every line through its foot in the plane. 

If a straight line is perpendicular to any two straight 
lines through its foot in a plane, it is perpendicular to the 
plane. 

The angle between two lines not in the same plane 
is the same as the angle between two intersecting lines 
parallel respectively to the given lines. 

195 



196 ANALYTIC GEOMETRY OF SPACE [Ch. 1, § 2 

The orthogonal projection of a point on a plane (or an 
axis) is the foot of the perpendicular from the point to 
the plane (or the axis). The projection of a portion 
of a line or curve on a plane (or an axis) is the locus 
of the projections of all its points. 

The angle which a line makes with a plane is the angle 
which it makes with its projection on the plane. 

The angle between two planes is measured by the angle 
between two lines, one in each plane, drawn perpendicular 
to their intersection at the same point. 

2. Rectangular coordinates. — In applying algebra to 
the geometry of space, we must first devise some method 
of representing the position of a point in space by 
numbers. 

Construct three mutually perpendicular planes, X-Y, 
Y-Z, and Z-X, dividing all space into eight compart- 
ments, called octants. These planes are spoken of as 
coordinate planes, their point of intersection, 0, as the 
origin, and their lines of intersection, OX, OY, and GZ, 
as coordinate axes. 

A point in space is located by means of its distances, 
AP, BP, and (XP, from the coordinate planes, measured 
parallel to the coordinate axes. The three numbers which 
represent these distances are called the rectangular coor- 
dinates of the point, and are always written in the order 
(x, y, z). 

We shall consider distances as positive when measured 
to the right, forward, or upward ; that is, parallel to 
OX, OY, and OZ. Distances measured in the opposite 
directions will then be negative, The octant 0-XYZ is 



Ch. 1, § 3] COORDINATE SYSTEMS. THE POINT 



197 



x 



«B 



called the first, and the others may be numbered in any 
convenient way. 

The position of any point (#, y, z) may be determined 
by taking on the axes the distances OZ, OM, and OJV, 
equal to these coordinates, 
and through the points X, 
M, iV, passing planes parallel 
to the coordinate planes, 
forming a rectangular par- 
allelopiped ; the point of 
intersection of these planes _ 
will be the point required. 

It is evident that rectan- 
gular coordinates in a plane 
is a special case of this more 
general system, in which one 

of the coordinates has become zero. We ought therefore 
to be able to reduce all of the formulas in three dimensions 
to the corresponding formulas in two dimensions by plac- 
ing z equal to zero. 



I 






Fig. 1. 



PROBLEMS 

1. Plot the following points : 

(5, 4, 3), (- 3, 4. 1), (- 3, - 1, 2), (2, - 3, 1), (1, 1, - 2), 
(-1, 4, -2), (-3, -2, -1), (4, -1, -2); (3, 4, 0), 
(-2, 0, 1), (0, -i, 3); (5, 0, 0), (0, 3, 0), (0, 0, -2). 

3. Distance between two points. — Let P x and P 2 be 

any two points in space, and through each of them 
pass three planes parallel to the coordinate planes, form- 
ing a rectangular parallelopiped. 



198 



ANALYTIC GEOMETRY OF SPACE 



[Cii. I, §4 



Since the square of the diagonal of a rectangular 

parallelopiped equals the 
sum of the squares of its 
edges, 




p x pf = pjtf + pa 



Eig. 2 



But P 1 B 1 = x 2 — x v 

and P\T X = z 2 — z v 



Hence P X P 2 = Vte - a>i) 2 + (2/2 - 2/1) 2 + (» 2 - zi) 2 . [1] 
The distance, p, of any point from the origin is 

p = V« 2 + y 2 + s 2 . [2] 



evidently 



4. To divide a line in any given ratio. — Let the point 
P divide the line P X P 2 so that 



P,P 



Project the line P X P 2 
on the X~r"-plane, form- 
ing the trapezoid P X C 2 , 
in which 1 P 1 = z v 
C 2 P 2 = s 2 , and CP = z. 
It will be noticed that 
this is the same figure 
used in Art. 13, Part I. 

Hence 

_ m 2 3i 4- niiz-2 t 
~ mi + nio 




Ch. I, §5] COORDINATE SYSTEMS. THE POINT 199 

If P X P 2 i s Projected on the other planes, we obtain in 
like manner 

= ma + mo* d = *M^. [3] 

If the line is bisected, these formulas become 

«=*!+*, 1,=^, and .=S + * [4] 



PROBLEMS 

1. Find the length of the line joining the two points 
(3, 2, — 1) and (4, — 2, 6) and the coordinates of the point 
which divide this line in the ratio 3 : — 2. 

2. Find the coordinates of the centre of gravity of the tri- 
angle whose vertices are (a^, y x , z x ) f (x 2 , y 2 , z 2 ), and (x s , y s , z 3 ). 

3. Prove that in any tetraedron the four lines joining the 
vertices with the centres of gravity of the opposite faces meet in 
a point, which is three-fourths of the distance from each vertex 
to the opposite face. (This point is the centre of gravity of 
the tetraedron.) 

4. Show that the centre of gravity of any tetraedron bisects 
each of the four lines joining the middle points of the opposite 
edges. 

5. Show that the straight lines which join the middle points 
of the opposite sides of any quadrilateral meet in a point and 
are bisected at that point. 

6. Show that the sum of the squares of the diagonals of any 
quadrilateral is twice the sum of the squares of the lines which , 
join the middle points of the opposite sides. 

5. Projection of a given line on a given axis. — It is 
required to find the projection on the axis OX of the line 
AB which makes an anode a with the axis. 



200 



ANALYTIC GEOMETRY OF SPACE 



[Ch. I, § 6 



Through A and B pass planes perpendicular to the axis, 
cutting it at A' and B' . Then A'B' will be the pro- 
jection of AB on OX. Through A draw the line AC 

parallel to OX. Then 
AC= A'B', and the angle 
CAB = a. In the right 
triangle ABC, 




AC 
AB' 


= cos a. 


ence 




A'B' = 


AB COS a 



Fig. 4. 



[5] 

That is, the projection of a line on an axis is equal to the 
length of the line multiplied by the cosine of the angle which 
the line makes with the axis. 

The projection, A'B', of a directed line, AB, is evidently 
a directed line. If a broken line AB, BC, CD is pro- 
jected on an axis, the algebraic sum of the projections 
of its parts will be the distance along the axis from the 
projection of A to the projection of D. The projection on 
any axis, then, of any closed path ABC... A in space, which 
is looked upon as generated by the movement of a point 
from A to B, B to C, etc., is zero. 

6. Polar coordinates. — Let OX, OY, OZ be a set of 

rectangular axes in space, and let P be any point. Draw 
OP. 

The position of P is evidently determined if we know 
its distance p from the origin, and the angles a, ft, and 7 
which OP makes with the coordinate axes. 

The distance p is called the radius vector of the point 



Ch. I, § G] COORDINATE SYSTEMS. THE POINT 



201 



P : «, /3, and 7, the direction angles of the line OP ; and 
the four quantities (p, a, /3, 7), the polar coordinates of the 
point. Cos a, cos/3, and cos 7 are called the direction 




cosines of the line OP, and may be represented by the 
letters ?, w, and n. 

Let X, M, and 2V~ be the projections of P on the axes. 
Then OL = x, OM= y and ON '= z ; and from right tri- 
angles we have 



jc—p cos a, 
y = p COS P, 

2! = p cos y. 



[6] 



These equations give the relations between the rec- 
tangular and polar coordinates of any point. Squaring 
and adding, we have 

x 2 -f y 2 -f z 2 = p 2 (cos 2 a -f- cos 2 ft + cos 2 7). 



But from [2], 
Hence 



P 2 = x 2 + y 2 + z 2 . 
cos 2 a + cos 2 p 4- cos 2 y = 1. 



[7] 



202 ANALYTIC GEOMETRY OF SPACE [Ch. I, § 6 

That is, the sum of the squares of the direction cosines 
of any line is unity. Hence the four quantities used as 
polar coordinates of a point are equivalent to only three 
independent conditions, as we should expect. 

We may always choose these coordinates so that they 
shall all be positive and so that the angles «, /3, 7 shall 
not be greater than 180°. 

Since any line parallel to OP makes the same angles 
with the axes, we may define the direction cosines of any 
line in space as the same as the direction cosines of a paral- 
lel through the origin. 

PROBLEMS 

1. Find the direction angles of a line equally inclined to 
the three axes. 

2. If /, m, and n are the direction cosines of a line, show 
that — I, — m, and — n are the direction cosines of the same 
line running in the opposite direction. 

3. Find the direction cosines of the line joining the origin 
to the point (2, 6, 2), and the projection of the line on each of 
the coordinate axes. 

4. Find the direction cosines of the line joining the points 
(2, 5, 1) and (3, 1, 8), and the projection of the line on each of 
the coordinate axes. 

5. Show that any three numbers are proportional to the 
direction cosines of some line. 

6. Find the direction cosines of a line which are propor- 
tional to the numbers 1, 2, 3. 

7. Show that the square of the distance between two points 
whose polar coordinates are (p 1? « 1? /? 1? y x ) and (p 2 , « 2 > fi-2> 72) is 

pi + pi — 2 p\p-2 (cos «x cos a 2 -f- cos /3 l cos /? 2 4- cos y 1 cos y 2 ). 

8. A line makes an angle of G0° with the X-axis, and 45° 
with the Y"-axis. What angle does it make with the Z-axis ? 



Ch. I, § 7] COORDINATE SYSTEMS. THE POINT 



203 



7. Spherical coordinates. — Let OX, OY, OZ be a set of 
rectangular axes in space, and let P be any point. Draw 
OP, and pass a plane through OZ and OP. The position 
of any point in space is determined, if we know the dis- 




Eig. 6. 



tance p from the origin to the point ; the angle 6 which 
the plane ZOP makes with the fixed plane ZOX; and the 
angle <j> which OP makes Avith OZ. 

The line OZ is called the polar axis, and the point the 
pole. About as a centre describe a sphere with OP as 
radius. The plane ZOP will intersect the sphere in a 
meridian circle. The angle 6 may be called the longitude 
of P, and the angle c/>, the colatitude. The distance p is 
called the radius vector and (p, <£, 0) are called the spherical 
coordinates of P. The arrows indicate the usual choice of 
positive direction. 



204 



ANALYTIC GEOMETRY OF SPACE 



[Ch. I, § 8 



Let the student show that the relations between rec- 
tangular and spherical coordinates are 
a? = p sin <|> cos 0, 

y - p sin <f> sin 9, [8] 

z = p cos <|>. 

Note. — Spherical coordinates have usually been called polar coordi- 
nates. But the application of the system described in Art. 6 is more 
nearly analogous to the uses of polar coordinates in two dimensions. 

8. Angle between two lines. — Let a v /3 V y v and a 2 , £ 2 , y 2 

be the direction aDgles of two lines, and let 6 be the angle 
between them. Draw parallels to these lines through the 
origin, and on each of these parallels take a point, as P x 
and P 9 . 




Fig. 7. 
Then by [1] 

i\p? = (x x - x 2 y + to - y 2 y + (ti - * 2 ) 2 > 

or by [6] = {p 1 cos « x — p 2 cos a 2 ) 2 -f- (p x cos ft x — p 2 cos /3 2 ) 2 

+ (/>iCOS7 1 - / d 2 cos 7 2 ) 2 , 
or by [7] = p^-\-p 2 — 2 p x p 2 (cos «x 1 cos a 2 + cos /3 X cos /5 2 

-J- cos 7 X cos7 2 ). 



Ch. I, § 8] COORDINATE SYSTEMS. THE POINT 205 

But by the law of the cosines 

IVV 2 = Pl 2 +p 2 2 - 2 PlP2 cos 6. 

Hence cos = cos ai cos a 2 + cos Pi cos p 2 + cos yi cos y 2 , [9] 

If the lines are perpendicular cos 6 = 0, and the condi- 
tion for perpendicularity is 

COS ai cos a 2 + cos pi cos p 2 + cos 71 cos -y 2 = 0. [10] 

If the lines are parallel, they must make the same angles 
with the axes, and the conditions for parallelism are - 

ai =aj, p! = po, and yi = Y2« [11] 

PROBLEMS 

1. Show that the three lines whose direction cosines are 

12 -3 -4. _4_ 12 _3_. nrif 1 3 -4 12 
13' 13' 13' 13' 13' 13' dlUA 13' 1 3' TT 

are mutually perpendicular. 

2. Show that (3, 30°, 60°, 90°), and (5, 30°, 90°, 60°) are 
possible polar coordinates of two points, and find the angle they 
subtend at the origin. 

3. Show that the conditions for parallelism are consistent 
with [9] when = 0°. 

4. Find the rectangular coordinates of the points in 
problem 2. 

5. Find the polar coordinates of the point (3, — 6, 2). 

6. Find the angle subtended at the point (1, 2, 3) by the 
points (2, 3, 4) and (o, 4, 3). 



206 



ANALYTIC GEOMETRY OF SPACE [Ch. I, § 10 



9. Transformation of coordinates. Parallel axes. — If 

the new axes are parallel to the old, and the coordinates 

of the new origin, re- 
ferred to the old axes, are 
(# , y , z ), the equations 
of transformation are 




-X' 



V' 



easily seen (see Fig. 8) 



^ to be 



X = X + X 1 , 

2/ = 2/o 4- y' 9 [12] 

z = So + z . 



10. Transformation of coordinates from one set of rec- 
tangular axes to another which has the same origin. — Let 
i a v Pv 7i)> ( a 2> && 72)' an( ^ ( a 3' $3* 73) ^ e the direction 
angles of OX', OY\ and 
OZ* with respect to the 
original axes. The coor- 
dinates (x, y, z) of any 
point P are the projec- 
tions of OP on OX, OY, 
and OZ. But the broken 
line made up of x f , y\ and 
z' extends from to P, 
and will therefore have 
the same projections on * 
the axes as OP. Hence 
(by Art. 5) 




Fig. 9. 



X = X' COS ai + y' COS a 2 + Z' COS a 3 , 
y = x' cos pi 4- y' cos (3 2 + z' cos p 3 , 
2 = x' cos 71 + 2/' cos -y- 4- s' cos 73. 



[13] 



Ch. I, § 10] COORDINATE SYSTEMS. THE POINT 207 

Let the student show that the transformation of coordi- 
nates cannot alter the degree of an equation. (See Art. 
50 Part I.) 

PROBLEMS 

1. What will be the direction cosines of OX, OY, and OZ 
referred to the new axes in Art. 10 ? 

2. What six relations hold between a l9 ft, y ]5 a 2 , ft, etc., 
from [7] ? 

3. What six relations hold between a lf ft, y 1} a 2 , ft, etc., 
from [10] ? 

4. Show that the twelve relations obtained in problems 2 
and 3 are equivalent to only six independent conditions. How 
many of the coefficients in equations [13] are independent ? 



CHAPTER II 
LOCI 

11. Equation of a locus. — If a point moves in space 
according to some law, it will generate some locus. As, 
for example, a point keeping at a fixed distance from a 
fixed point will generate the surface of a sphere. If we 
can translate the statement of the law into an algebraic 
relation between the coordinates of the points which satisfy 
the law, we shall have, as in plane analytic geometry, an 
equation which can be used to represent the locus. In 
the above example, if the origin is at the centre, the equa- 
tion of the surface- will be x 2 + y 2 + z 2 = r 2 ; for this states 
that the point (x, y, 2), which satisfies it, must remain at 
the distance r from the origin. 

The planes parallel to the coordinate planes are evi- 
dently represented by x = Jc v y = Jc 2 , and z = k s ; for these 
equations state that the points which satisfy them are at a 
fixed distance from the coordinate planes. 

PROBLEMS 

1 . What are the equations of the coordinate planes ? 

2. What are the equations of the planes bisecting the angles 
between the X- T and I -Z-planes ? Between the Y-Z and 
Z-X-planes ? 

3. AYhat equation must be satisfied by the coordinates of a 
point which remains at a distance of 5 units from the X-axis ? 
5 units from the F-axis ? What is the locus in each case ? 

208 



Ch. II, §§ 12. 13] LOCI 209 

4. Find the equation of the locus of a point which is 5 
units from the point (3, 2, 5). 

5. What equations must be satisfied by the coordinates of 
a point which is equidistant from the three points (1, 3, 8), 
(-6, -4, 2), and (3, 2,1)? 

12. Cylindrical surfaces. — If a cylindrical surface is 
formed by the movement of a line, which remains parallel 
to one of the axes, while moving along a directing curve 
in the plane of the remaining axes, its equation in three 
dimensions will be the same as the equation in two dimen- 
sions of the directing curve, and will contain only two 
variables. For, suppose the line remains parallel to the 
Z-axis and the directing curve lies in the JT-^F-plane ; then, 
for any position of the line, the relation between the x and 
y coordinates of any point on it will be the same as the 
relation between the x and y coordinates of the point where 
the line touches the directing curve, while the z coordi- 
nate may have any value whatever. The equation in x 
and y of the directing curve is, therefore, the only necessary 
relation between the coordinates of any point on the sur- 
face, and as it is not satisfied by any point not on the 
surface, it is (when interpreted as an equation in three 
dimensions) the equation of the surface. 

In a similar manner, it may be shown that the equations 
of cylindrical surfaces, whose elements are parallel to the 
X-axis, contain only y and z ; parallel to the Z"-axis, only 
x and z. 

13. Surfaces of revolution. — Surfaces generated by the 
revolution of a plane curve about one of the coordinate 
axes form another class of surfaces whose equations can 
be determined easily. 



210 



ANALYTIC GEOIY1ETRY OF SPACE [Ch. II, § 13 



For example, let it be required to determine the equation 
of the surface generated by the revolution of the ellipse 
1 about the X-axis. Let P' (V, y 1 ', z') be any 
z 



a^b 2 




Fig. 10. 



point on the surface, and through P' pass a plane perpen- 
dicular to the X-axis. The section of the surface made 
by this plane is evidently a circle. Hence LP' = LK. 

But LP' = VV 2 + z' 2 and OL = x[. 

The coordinates of K in the X-!F-plane are, therefore, 
x' and Vy' 2 + z' 2 , and since K is a point on the ellipse 

x 2 v 2 

-^ + ^ = 1, these coordinates must satisfy that equation, 



or 



- + € - 



! + s' 5 



b 2 



1. 



Dropping primes, we have as the equation of an ellipsoid 
of revolution about the X-axis, 



x 2 



-+£- + — = 

?^b 2 ^b 2 



1. 



Ch. II, § 14] LOCI 211 

A general rule for finding the equation of a surface of 
revolution, formed by revolving a plane curve about one 
of the coordinate axes, may be stated thus : Replace in the 
equation of the plane curve the coordinate perpendicular to 
the axis of revolution by the square root of the sum of the 
ares of itself and of the third coordinate. 



PROBLEMS 

1. Find the equation of the surface generated by a line 
moving parallel to the Z-axis along 

(a) the ellipse -5 + ^ = 1, 
v ; a- b- 

(b) the parabola y 2 = 2 mx, 

(c) the line x + 3 y = 6. 

2. What is the equation of a circular cylinder whose axis is 
parallel to the F-axis and passes through the point (3, 0, 5), 
and whose radius is 5. 

3. Find the equation of the surface of revolution, formed by 
revolving about the X-axis 

(a) the line y = 4, (a cylinder) 

(b) the line x = y, (a cone) 

(c) the circle x 2 + y- — r 2 , (a sphere) 

(cl) the parabola y 2 = 2 mx, (a paraboloid of revolution). 

4. Obtain the equations of the hyperboloids of revolution 
formed by revolving the hyperbola about (a) its transverse 
axis; (b) its conjugate axis. 

14. Locus of an equation. — Again, as in plane analytic 
geometry, an equation between x, y, and z expresses a 
necessary relation between the coordinates of every point 
which satisfies it, and hence cannot be satisfied by points 
taken at random in space. It is easy to see that the 



212 ANALYTIC GEOMETRY OF SPACE [Ch. II, § 14 

points which satisfy it may be taken as near to each other 
as we please. Moreover, any such equation represents a 
surface of some kind, as we shall now prove. 

Let f(x, y, z) = be an equation of any degree between 
#, ?/, and z. If we substitute x = k (any constant), the 
resulting equation, f(y, z) = 0, must represent the rela- 
tion between y and z for all points of the locus for which 
x = &, or which lie in a plane at distance k from the 
F-Z-plane. But since the locus of /(«/, z) = lies wholly 
in this plane, it is a plane curve. Hence the intersection 
of any plane parallel to the F-Z-plane with the locus of 
f(x, y, z ) = is a plane curve. This can be proved in 
like manner for all planes parallel to the X-Y and X-Z- 
planes. If the axes are revolved through any angle, the 
equation of the locus will be of the same general form 
and every plane parallel to the new axes will cut it in a 
plane curve. Hence all planes cut the locus in a plane 
curve, and the locus is therefore a surface. 

If, in particular, the equation is of the first degree, its 
intersection with any of these planes will be a straight 
line. An equation of the first degree therefore alivays 
represents a plane. 

If an equation does not contain a term in z, the relation 
between x and y will not be changed by a change in z. 
The sections of the locus parallel to the X-Z-plane are 
therefore all alike, and the locus is a cylindrical surface, 
having all its elements parallel to the Z-axis. In like 
manner, if an equation does not contain a term in y, it 
represents a cylindrical surface parallel to the Z"-axis ; 
if it contains no term in x, a cylindrical surface parallel 
to the X-axis. 



Ch. II, § 14] LOCI 213 

If in particular the equation is of the first degree, the 
surface becomes a plane parallel to one of the axes. 

If two equations are simultaneously satisfied by the 
coordinates of points on a locus, that locus must consist 
of the points common to the loci of the two equations. 
Hence two equations of the form f x (x, y, z) = and 
/ 2 (#, y, z) = 0, taken together, represent a curve in sjiace, 
the intersection of the surfaces which they represent. 

In particular, if these two equations are of the first 
degree, this locus will be the intersection of the two 
planes which they represent. Hence two equations of the 
first degree, use J simultaneously, represent a straight line. 

Three equations used simultaneously are satisfied by the 
coordinates of a finite number of points only, — the points 
of intersection of the curve represented by two of the equa- 
tions with the surface represented by the third. 

The curves of intersection of any surface with the 
coordinate planes are called the traces of the surface. 
Their equations may be found from the equation of the 
surface by placing each of the coordinates in turn equal 
to zero. 

The general method of determining the form of the 
surface represented by any given equation will be taken 
up in the chapter on quadric surfaces. 

PROBLEMS 
1. What surface is represented by the equations ? 
(a) x = y, (d) x> + tf = 2o, 

0) V = z, (e) st? + ir + z 2 = 25, 

(c) x-y=5, (/) z 2 -2y = 0. 



214 ANALYTIC GEOMETRY OF SPACE [Ch. II, § 14 

2. Obtain the traces on each of the coordinate planes of the 
loci of the following equations, and from these traces deter- 
mine roughly the nature of the surface : 

(a) x 2 + y 2 = 9, (d) y 2 = ±z, 

(6) x -y + 2z = 10, (e) | + J + g = l, 

3. What is the equation of the surface generated by the 
revolution of the hyperbola xy == k about the X-axis. 

4. What is the position of a line whose equations are 
x + 3y = 10 and 3a- 4^ = 8? 

5. The equations of any two surfaces may be represented 
by U=0 and V= 0, where U and Fare abbreviations for 
algebraic expressions of any degree in x, y, and z. Show 
that lU+kV=0 will represent a surface which passes 
through all the points common to the loci of U=0 and 
V=0, and which meets neither of these surfaces at any 
other points. Show also that the locus of UV= will 
consist of the loci of U= and V= 0. 



CHAPTER III 



THE PLANE 

15. Normal form of the equation of a plane. — Let ON 

be the normal to the plane (a straight line of indefinite 
extent perpendicular to the plane), and let «, /3, and 7 be 
the angles which this normal makes with the axes. Let 
p be the perpendicular distance OK from the origin to 




Fig. 11. 



the plane, measured along the normal. Let P(x,y,z) 
be any point in the plane. The line Pif will be perpen- 
dicular to ON, and the projection of OP on ON will be 
OK or p. But the projection of OP on ON is the same 
as the projection on ON of the broken line OL, L C, CP, 

215 



216 ANALYTIC GEOMETRY OF SPACE [Ch. Ill, § 16 

or x, y, z. From [5] the projection of OL on ON is 
x cos a ; of LC y cos /3 ; of (7P, 2 cos 7. 

Hence a? cos a + ycosp + scos-y — p = 0. [1*^] 

This is called the normal form of the equation of a 
plane. 

The distance p is measured from the origin to the 
plane, and is positive or negative according as it runs 
in the positive or negative direction of the normal. It 
is usually possible to choose the direction from the origin 
to the plane as the positive direction of the normal,, so 
that p will usually be a positive number. 

The angles a, /3, and 7 are measured from the positive 
directions of the axes to the positive direction of the 
normal. 

16. Reduction of the general equation Ax + By + Cz 
-f D = to the normal form. — It has been shown in the 
previous chapter that every equation of the first degree 
represents a plane. Let the general equation of the first 
degree, Ax + By + Cz + D = 0, be the equation of a plane, 
and let x cos a + y cos fi + z cos 7 — p = be the equation 
of the same plane in the normal form. Then, since the 
two equations represent the same plane, they can differ 
only by a common factor. Then JcA = cos a, kB = cos /3, 

and 7cC= cos 7. Hence k = - , and the 

,. ± V^L 2 + B 2 + C 2 

equation T ^ 



z&+ V + - 



±VA* + B 2 +C- 2 ±VA 2 + B 2 +C 2 ±VA 2 + B 2 + C 1 

+ n =0 [15] 



Ch. Ill, § 18] THE PLANE 217 

is in the normal form. If we wish to keep p positive, it 
is necessary to choose the sign of the radical opposite to 
the sign of D. Then the coefficient of x is cos a, etc. 

17. Equation of a plane in terms of its intercepts. — If 
the intercepts of a plane on the axes are #, 5, and c, the 
coordinates of the points where it cuts the axes are 
(a, 0, 0), (0, 5, 0), and (0, 0, c). If these coordinates 
are substituted successively in the general equation 

Ax + By + Cz + D = 0, 
we have a = — -, b— — -, and c = -■ 

But the general equation may be written in the form 
x y z 1 

ABC 
From this we have, by substitution, 

- + ? + - = l [16] 

a b c L J 

as the equation of a plane in terms of its intercepts. 

18. Distance of a point from a plane. — Let it be re- 
quired to find the distance of the point P x from the plane 
HK, when the equation of ffKis given in the form 

x cos a -f- y cos ft + z cos <y — ^> = 0. 

Pass a plane BS through P v parallel to HK. Its 
equation will be 

x cos u + y cos ft -f- z cos 7 — jpj = 0, 




218 ANALYTIC GEOMETRY OF SPACE [Ch. Ill, § 18 

where p 1 ^ can be either positive or negative, since it 

is the distance from the 
origin to the plane RS, 
measured along the nor- 
mal to UK. The coordi- 
nates of P x must satisfy 
the equation of US. 
Hence 

x 1 cos a -f y l cos ft 

+ z x cos 7 = p v 

Y Now, wherever P 1 may 

ElG - 12 - lie, 

MP X = N^ = 0^ - ON=p 1 -p 

= x 1 cos a + y 1 cos /3 + z x cos 7 — p. 

If the equation is given in the form 

Ax + By + <7z + D = 0, 

M1>i = -Abi + #2/1 + C«i + 2> f ri7 -. 

± y/A* + B 2 +C 2 L J 

where the sign of the radical is chosen opposite to that of 
D. The distance MP X is positive when the point and 
the origin are on opposite sides of the plane ; negative 
when they are on the same side of the plane. 

PROBLEMS 

1. Given the plane 3x — 8y + z = 12, find . 

(a) the direction cosines of a normal, 

(b) its distance from the origin, 

(c) its distance from the point (3, — 2, 6), 

(d) its intercepts. 



Ch. Ill, § 19] THE PLANE 219 

2. Find the equation of a plane, if the foot of the perpen- 
dicular from the origin on it is the point (3, 1, — 5). 

3. On which side of the plane 7a; + 4y = 5 is the point 
(0, 7, 3) ? How is this plane situated ? What are its traces 
on the coordinate planes ? 

4. Show that the three planes 2x + 5y + 3z = 0, x — y 
+ 42 = 2, and 1 y — 5 2 + 4 = intersect in a straight line. 

5. Find the equation of the plane which bisects the 
angle between the two planes A x x + B x y + C x z + D y = and 
A& + B$ + C& + D, = 0. 

6. Find the equation of a plane through the origin and 
the line of intersection of the planes x + 3 y — 4 z = 10 and 
5?/ — 62 + 3 = 0. (See problem 5, page 214.) 

19. The angle between two planes. — The angle between 
two planes is easily seen to be equal to the angle between 
their normals. 

If the two planes are 

x cos a x + y cos /3 1 + z cos y 1 —p 1 = 0, 
and x cos « 2 + y cos fi 2 + z cos y 2 — /> 2 = 0, 

the angle between them is given by 

COS0 = COS ai COSa.2 + cos pi cos p 2 + cos vi cos 72. [18] 

If the two planes are A x x + B x y + C x z + D x = 0, 
and A 2 x + B 2 y + 6 7 2 2 + D 2 = 0, 

the angle between them is given by 

cos0 = A 1 M + B 1 B % +C 1 Ct r 191 

± y/Ai* + B^ + C1 2 ' VA2 2 + # 2 2 + C 2 ' 2 

If the sign of the first radical is chosen opposite to the 
sign of D v and the sign of the second opposite to the sign 



220 ANALYTIC GEOMETRY OF SPACE [Ch. Ill, § 21 

of 2> 2 , 6 will be the angle between the positive directions 
of normals to the planes. 

20. Perpendicular and parallel planes. — If two planes 
are perpendicular, cos = 0, 
and AiA 2 + BtBi + dC 2 = 0. [20] 

If two planes are parallel, the direction cosines of their 
normals must be equal, 

or ^l = ^ 



V^ 2 + B? 4- C* VA 2 2 + B 2 2 + <7 2 2 
B* B () 



and 



V^ 2 4- Bf 4- C* -J A} 4- B* 4- 6 2 2 
Y\ ^2 



V^ 2 + B? 4- C\ 2 V^ 2 2 4- B 2 2 + <? 2 2 
The conditions for parallelism are, therefore, 



A 1 = B 1= C lm 

A2 B2 C/2 



[21] 



Notice that two planes will be perpendicular when a 
single condition is satisfied; but that two conditions must 
be satisfied if the two planes are to be parallel. 

21. The equation of a plane satisfying three conditions. — 

The general equation of a plane, Ax + By 4- Cz 4- D = 0, 
contains three independent coefficients, and therefore three 
independent relations between the coefficients will deter- 
mine the plane. For any three such conditions will give 
three equations between the four coefficients, from which 
three of the coefficients can be determined in terms of the 
fourth. If we substitute these values in the general 
equation, and divide by the fourth coefficient, the equa- 
tion is completely determined. 



Ch. Ill, § 21] THE PLANE 221 

The coordinates of three points are three conditions 
from which three such equations can be obtained; for 
these sets of coordinates must each satisfy the general 
equation. If the three points happen to lie on a line, the 
equations for determining the coefficients will not be inde- 
pendent, and the plane will not be determined. 

Again, a plane can be determined which shall pass 
through two points and also be perpendicular to a given 
plane ; for the substitution of the coordinates of the two 
points will give two equations between the coefficients, 
and the condition for perpendicularity [20] will give a 
third. If, however, the two points lie on the same normal 
to the plane, the solution will be indeterminate, since the 
conditions will not be independent. Again, a plane can be 
determined which shall pass through one point and also 
be parallel to a given plane ; for the substitution of the 
coordinates of the point gives one equation between the 
coefficients, and the conditions for parallelism [21] give 
a second and third. 

But here there is a simpler method ; if the equation of 
the plane is Ax -f- By + Cz + D = 0, any plane parallel 
to it may be written in the form Ax +By -f Cz +D 1 = 0, 
since the conditions for parallelism are satisfied. We can 
determine D l from the fact that the coordinates of the 
point must satisfy the equation. 

PROBLEMS 
1. Find the equation of a plane through the points 

(a) (4, 2, 1), (- 1, - 2, 2), (0, 4, - 5), 

(b) (-1,-1, -1), (3,2, -2), (2,0,0). 



222 ANALYTIC GEOMETRY OF SPACE [Ch. Ill, § 21 

Find the intercepts of these planes on the axes and their 
distances from the origin. 

2. Find the equation of a plane through the points (2, 1, —1) 
and (1, 1, 2), and perpendicular to the plane lx + 4ty — 4z 
- 36 = 0. 

3. Find the equation of a plane through the points (2, 0, —1) 
and(l, — 6, 1), and perpendicular to the plane ox+3y— z— 4=0. 

4. Find the equation of a plane through the point (2, 1, — 1) 
and parallel to the plane 7 x + 4 y — 4 2 + 36 = 0. 

5. Find the equation of a plane which bisects the line join- 
ing the two points (6, 4, 1) and (2, 4, —1), and is perpendicular 
to that line. 

6. Find the equation of a plane which passes through the 
origin and is perpendicular to the two planes 2x— 4y-f-3z=12 
and 7x + 2y + z = 0. 

7. Prove that the six planes, each containing one edge of a 
tetraedron and bisecting the opposite edge, meet in a point. 

Note. — The coordinates of the point of intersection of three planes 
may be found by solving the three equations simultaneously. 

8. Prove that the six planes, each passing through the mid- 
dle point of one edge of a tetraedron and being perpendicular 
to the opposite edge, meet in a point. 



CHAPTER IV 

THE STRAIGHT LINE 

22. Equations. — We have seen that, if a point moves 
in space in such a way as to satisfy at the same time two 
equations of the first degree, the locus which is generated 
is the line of intersection of their planes. Then the two 
equations 

A^x + B x y + G x z + D 1 = 0, 

and A 2 x + B 2 y + C 2 z + D 2 = 

will in general represent a line, the only exception being 

ABC 

when — J = — l = — l, and the planes are parallel. 
A 2 B 2 C 2 

But the line may be determined by any pair of planes 
which pass through it, and it is convenient to pick out 
those planes which have the simplest form. The equa- 
tion of any plane through the line can be written in the 
form 
A x x + B x y + C\z + D 1 +k (A 2 x + B 2 y + C 2 z + 2> 2 ) = 0. 

When none of the coefficients A v B v etc., are zero, it 
will always be possible to choose k in such a way as to 
eliminate y and reduce the equation to the form x = mz-\-a. 
Again, Jc may be so chosen as to eliminate x and reduce 
the equation to the form y = nz + b. Then the equations 

x = mz + a, 

and y = nz + b, 

223 



224 ANALYTIC GEOMETRY OF SPACE [Ch. IV, § 22 

each determine a plane through the line, and hence may 
be used as the equations of the line. These planes are 
seen to be the projecting planes of the line, perpendicular 
to the X-Z and F-Z-planes. The equations of any two of 
the three projecting planes may be chosen as the equations 
of the line. 

In practice, to reduce the equations of a line to their 
simplest form, we simply eliminate one of the variables 
and then another from the two equations. Indeed, it is 
evident algebraically that any set of values which satisfy 
a pair of equations must also satisfy any equation which 
can be deduced from them. 

If some of the coefficients A v B v etc., are zero, it will 
always be possible by elimination to reduce the equations 
to one of the three forms 

x = mz + a, y = qx + c, x = e, 

or 
y = nz + 5, z = <2, z =f. 

The first form includes all lines not parallel to the X-Y- 
plane ; the second, lines parallel to the X-F'-plane, but 
not parallel to the 3^-axis ; the third, lines parallel to the 
Y-axis. 

PROBLEMS 

1. Write the equations of each of the coordinate axes. 

2. Write the most general form of the equations of a line in 
each of the coordinate planes ; parallel to each of the coordi- 
nate planes ; parallel to each of the coordinate axes. 

3. Show how to find the points where a given line pierces 
the coordinate planes, and by this means plot the lines in 
problem 4. 



Ch. IV. § 23] 



THE STRAIGHT LINE 



225 



4. Reduce these equations to their simplest forms : 



(a) 2^-3?/-h z-6 = 0, 

x + ^-33-1 = 0. 

(c) 2a+-4y + 3z + 6 = J 

3a; + 6# + 2z — 1 = 0. 
(e) 2x'-3y- 2 + 2 = 0, 

4aj-6y + 3z-l = 0. 



(6) 2x + 3y- 6^-12 = 0, 
4a;- # + 122 + 4 = 0. 

(d) 4*/ + 3 2+ 1 = 0, 
3^-22-12 = 0. 

(/) 42/ + 32- 2 = 0, 

2y- 2 + 4 = 0. 



5. Find the equations of the line of intersection of the plane 
2 as — 3y-\-z — 6 = with the coordinate planes. 

23. The equations of a line in terms of its direction 
cosines and the coordinates of a point through which it 
passes. — Let a, /3, and 7 be the direction angles of the 

\z / 




Fig. 13. c 

line and P 1 a point through which it passes. Let P be 
any point on the line. Then from the figure 

x — x 1 = P X P cos a, 

2/ -i/ l = P l P cos ft 

2 — 2 X = -P^P COS 7. 



226 ANALYTIC GEOMETRY OF SPACE [Ch. IV, § 24 

Solving these for P 2 P, and equating the values, we have 

OC-OCi _y-Pl_Z-2l 



COS a COS0 COSX 



PROBLEMS 



[22] 



1 . What form will these equations take when a — 90° ? 
when a = 90°, and /? = 90° ? 

2. Find the equations of a line through the point (— 1, 
2, -3) if 

(a) a = 60°, £ = 60°, y = 45°; 

(6) a = 120°, = 60°, y = 135°; 

(c) cos « = i V3, cos /? = 1, COS y = 0. 

Show that the given values are possible in each case and plot 
the line. 

3. Find the equations of a line through the origin, equally 
inclined to the axes. 

24. Given the equations of a line, to find its direction 
cosines. — The method is best shown by an example. Let 
the equations of a line, reduced to their simplest form, be 

x = dz — 6, and y = 2 z + 3, 

x + 6 y - 3 2-0 



Let x-^^y-j^ z-z, 

cos a cos p cos 7 

be the equation of the same line. These equations are of 
the same form and, since they represent the same line, 



v x = — 6, y 1 = 3, and z 1 = 0, 



Ch. IV, § 24] THE STRAIGHT LINE 227 

and the denominators, 5, 2, and 1, are proportional to cos a, 
cos /3, and cos 7. They can be made identical with them 
by multiplying by a suitable factor R. 

Then cos a = 5 22, cos ft = 2 22, and cos y= JR. 

Then by [7] 25 22 2 + 4 E 2 + 22 2 = 1, and 22 = — • 

VM 



u 5^2 1 
Hence cos a = , cos p = — — , cos 7 = , 

V30 V30 V30 

and the equation can be written in the form 
x+6 _y -3 _z-0 



V30 V30 V30 



PROBLEMS 



1. Show that, if the equations of a line can be written in 
the form x = mz + a, and y = nz + b, they may be changed 
into the form 

x — a y — b z 



Vm 2 -\- n 2 -\-l Vm 2 + >r + 1 Vm 2 + >r + 1 

^ = cos a, —- = ^ === ^ =z == cos /?, 

Vm 2 + ?r + 1 Vm 2 -+- >i 2 + 1 

and — — - = cos y. 

Vm 2 + n 2 + 1 

2. What form will the equations take, if their simplest 
forms are 

2/ = gx + c, cc = e, 

2f=^ Z=f? 



228 ANALYTIC GEOMETRY OF SPACE [Ch. IV, § 26 

3. Find the direction cosines of the lines whose equations are 

(a) 2x + 3y-2z -13 = 0, 

3 x + 6 y - 2 z - 24 = 0. 

(b) 2x + 2y-3z- 2 = 0, 

4 x — y — z — 6 = 0. 

(c) 2a + 4?/ + 3z + 6 = 0, 
3 a- + 6 y + 2>z- 1 = 0. 

(d) 4^ + 3^4- 1=0, 
3 y _2 2 -12= 0. 

25. Equations of a line through two points. — Let (x v 
y v z^) and (# 2 , «/ 2 , 3 2 ) be the two points. The equation 
of any line through the first point is (by [22]), 

x - x \ _ y - y\ _ z - z \ 



cos a cos ft cos y 
If the second point lies on this line, 

X 2 ~ x i _ ^2 ~ V\ _ Z 2 ~ z \ ± 
cos a ' cos ft " cos 7 

Dividing, we have, as the equations of a line through the 

two points, 

x - agi _ y -y\ = g-gi t [-23] 

PROBLEMS 

1. Establish equation [23] from an independent figure 
without using equation [22]. 

2. Discuss the special cases of [23], when x 2 = x 1 , y 2 — y\, 
or z 2 = z x . 



Ch. IV, § 25] THE STRAIGHT LINE 229 

3. Find the equations of a line passing through the points 

(a) (0, 0, - 2) and (3, - 1, 0), 

(b) (-1,3,2) and (2, -2,4), 

(c) (2, - 3, 1) and (2, - 3, - 1). 

4. Find the equations of the line joining the origin with 
the intersection of the planes 

x + 4 y + 2 z = 0, 

5. Are the three points (1, - 1, 2), (2, 3, - 1), and (3, 2, 2) 
in a straight line ? 

6. Show that the two lines 

x—2=2y-6=S z, 

and 4 a; — 11 = 4 y — 13 = 3 z 

meet in a point, and that the equation of the plane in which 
they lie is 

2 x - 6 y + 3 z + 14 = 0. 

7. Show that the line 4.r = 3?/ = — 2 is perpendicular to 
the line 3 x = — y = — 4 z. 

8. Find the point of intersection of the line 

2 x- 4=3y+l =2 +6 
with the plane ic -f- 6 2/ — 3 2 = 16. 

9. What is the equation of the plane determined by the 
point (3, 2, — 1) and the line 2x — o = oy + 1 = z? 



CHAPTER V 
QUADRIC SURFACES 

26. The sphere. — A sphere may be defined as the locus 
of a point whose distance from a fixed point is constant. 

If (# , y , z ) is the centre and r the radius, the equa- 
tion of the sphere is evidently 

(a? - xoT + (y - i/o) 2 + (* - ^o) 2 = r\ [24] 

If the centre is at the origin, the equation becomes 

K 2 + 2/ 2 + s 2 = r 2 . [25] 

Expanding [24], we see that the equation of every 
sphere is of the form 

as 2 + y 2 + z 2 + Gx + mj + Iz + K=:0, [26] 

where 

_a _H I 

x o — <£-> Vo— 2 ' z ° ~~ 2' 



and r = l V# 2 + JT 2 + i 2 - 4if. 

Every equation in the form of [26] will therefore repre- 
sent a sphere, 

real, if G 2 + IT 2 + 2 2 - 4 iT> 0, 

null, if £ 2 + H 2 + I 2 - 4 K= 0, 
imaginary, if # 2 + H 2 + 7 2 - 4 ^T< 0. 
Comparing [26] with the general equation of the second 
degree, 

Ax 2 + By 2 + Cz 2 + Dyz + Ezx + Fxy + Gx + Hy + Iz + K=0, 

230 



Ch. V, § 26] QUADRIC SURFACES 231 

we see that the general equation will represent a sphere, if 

D = U=F=0, and A = B = C. 

A sphere may, in general, be passed through any four 
points ; for the substitution of their coordinates in [26] 
will give four equations which will, in general, determine 
£, H, 7, and K. 

PROBLEMS 

1. Find the equation of a sphere with 

(a) centre at (o, — 2, 3), radius equal to 1. 

(b) centre at (2, — 3, — 6), passing through the origin. 

(c) centre on the Z-axis, radius a, passing through the 
origin. 

2. Find the centre and radius of each of the following 
spheres, when real : 

(a) x 2 + f + z 2 - 2x + 6y - Sz + 22 = 0. 

(b) x 2 + if + z 2 + 10a - 4 y + 2z + 5 = 0. 

(c) 3x 2 + 3y- + 3z 2 + 12a + 12 y + 18 2 + 3 = 0. 
(cZ) a 2 + f' + r + 6^= 0. 

(e) a 2 -f y 2 + z 2 + 4 x + ?/ + 5 2 + 21 = 0. 

3. Find the equation of the sphere passiug through the four 
points, 

(a) (2, 5, 14), (2, 10, 11), (2, 5, - 14), (2, - 10, - 11), 
(6) (0,0,0), (2,8,0), (5,0,15), (-3,8,1). 

4. Find the equation of a sphere passing through the origin 
and concentric with the sphere through the points (7, 7, 8), 
(- 1, - 5, - 8), (- 5, 7, - 6), (3, - 5, 10). 

5. Find the equation of a sphere with its centre at the 
origin and touching the sphere 

x 2 + y 2 + z 2 - 8 x - 6 y + 24 z + 48 = 0. 



232 ANALYTIC GEOMETKY OF SPACE [Ch. V, § 27 

6. Show that the equation of the sphere whose diameter is 

the line joining the points (x\. y lr z 2 ) and (x 2 , y 2 , z 2 ) may be put 
in the form 

(x - .i\) (x - .ro) + (y - y^ (y - y 2 ) +(z — z y ) (z - z,) = 0. 

7. Show that the equation x- + y' 2 + z 2 = y* 2 will have the 
same form, if the axes are turned through any angle without 
changing the origin. 

27. Conicoids. — Any surface whose equation is of the 
second degree in x, y, and z is called a quadric surface 
or conicoid. The sphere is a special case of such a 
surface. 

It is possible, by suitable transformation of coordinates, 
to reduce the general equation of the second degree in 
x, y. and z to one or other of these two forms, 

(1) Ax 2 + By 2 + Cz 2 = Z>, 

(2) Ax 2 + By 2 = Cz, 

where A, B, (7, and J) may be any quantities, positive, 
negative, or zero. But for our present discussion, let 
neither A, B, nor C vanish. 

The locus of equation (1) is evidently symmetrical 
with respect to each of the coordinate planes, and hence 
with respect to the origin. Such surfaces are therefore 
called central quadrics. 

If D =£ 0, equation (1) may be written in the form 

±^±g±g=l. [27] 

a- b 2 c 2 L J 

If D = 0, it may be written 



±£±S = «- [28] 






Ch. V, § 28] QUADRIC SURFACES 233 

Non-central quadrics are included under equation (2). 
It may be written in the form 



a 2 b 2 



-- 2 cz. [29] 



We shall now investigate the forms of the surfaces 
represented by these equations. 

sy*lk nj& /yii 

28. The ellipsoid. — + f- + — =1. — The surface is 
a 1 b l cr 

symmetrical with respect to each of the coordinate planes. 
Its intercepts on the X, Y", and Z-axes are ± a, ± b, and 
± c. The section of the surface made by the X-Y- plane 

is obtained by putting 2 = 0, and its equation is -— +^ = 1, 

a 2 b 2 

which represents an ellipse with semi-axes a and b. The 
section made by a plane parallel to this coordinate plane 
is found by putting z = z v This gives 

2! + 2? = i_£l, or ± | y 2 -i 

- " / ,q-%\i-$ • 

which represents an ellipse, in the plane z = z v with 

semi-axes ayl ^ and 6-yl — -L the centre lying on 

the Z-axis. 

As z 1 increases numerically from to ± <?, the section 
diminishes in size, until when z t — c it shrinks to a null 
ellipse, the single point (0, 0, c). As z x increases nu- 
merically beyond ± c, the section becomes imaginary ; 
hence the surface does not extend beyond the planes 
z — c and z = — e. 

Similarly, the sections made by the coordinate planes 



234 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 28 

Y-Z and Z-X and by planes parallel to them will be 
found to be ellipses with centres along the X and Y- 
z 



Fig. 14. 



axes respectively, and diminishing in size as the cutting 
plane moves off from the origin to the points where the 
surface cuts the axes. 




Fig. 15. 

rpu riili <yli 

If a = b, the equation becomes — + £- + — = 1. 

a A a z <r 

Sections made by planes parallel to the JL-y-plane will 
now be circles, whose centres lie along the Z-axis, and 



Ch. V, § 29] QUADRIC SURFACES 235 

the surface is an ellipsoid of revolution about the Z-axis. 

Similarly, ^+^+^=1 and -_ + 2L + fL = i repre- 
cr o z ¥ a L b l a 1 

sent ellipsoids of revolution about the X and Z"-axes 

respectively. 

If a = b = c, the ellipsoid becomes the sphere, 

x 2 -f- y 2 + z 2 = a 2 . 

The equation '---\-^ r --\- /i - = — 1 is not satisfied by any 
a 1 ¥ c l 

real values of .r. y, and 2 ; it may be said to represent an 

imaginary ellipsoid. 



x- 



j/2 *2 



9 = 1.-The 
a 1 b l c l 

intercepts on the X and F-axes are ± a and ± b, but the 

surface does not cut the Z-axis. 

The section of the surface made by the X-Z-plane is the 

ellipse — + y~ — 1 ; the section made by the F-Z-plane is 

Cf ^22 
the hyperbola ^- = 1, with its transverse axis, 2 b, 

along the I r -axis : the section made by the Z-X-plane is the 

hyperbola — — — = 1, with its transverse axis, 2 #, along 
the X-axis. 

The sections parallel to the X-F-plane will be ellipses, 
with their centres on the Z-axis ; the size of the ellipses 
will increase without limit as the cutting plane recedes 
from the X-I^-plane in either direction. 

We have now sufficient information to draw the figure. 

It is instructive, however, to investigate the plane sec- 
tions parallel to the other two coordinate planes. 



236 ANALYTIC GEOMETRY OF SPACE [Ch, V, § 29 

The section made by the plane x — x v parallel to the 
y-Z-plane, may be written 



y z 



K 1 -*) K 1 -^) 



= 1. 



If x x < a, this represents an hyperbola with its transverse 
axis parallel to the ]F-axis. As x x increases from to #, the 




Fig. 16. 

semi-axes both approach zero, and the hyperbola approaches 
a pair of intersecting lines. When x x — a, the section is 
the pair of straight lines, %- — — = 0, in the plane x — a, 

intersecting on the X-axis. When x x > a, the equation 
again represents an hyperbola, but the transverse axis is 
now parallel to the Z-axis. As x x increases, the semi-axes 
increase without limit. 



Ch. V, § 29] QUADRIC SURFACES 237 

Similarly, the sections made by planes parallel to the 
Z-X-jAsaie will be found to be hyperbolas, the transverse 
axis being parallel to the X-axis, when the distance of the 
cutting plane is less than 6, and parallel to the Z-axis, when 
the cutting plane is beyond y = b ; the transition from one 
set of hyperbolas to the other being a pair of intersecting 
lines in the plane y = b. 




Fig. 17. 

The surface is called the hyperboloid of one sheet, or the 
unparted hyperboloid, extending along the Z-axis. The 

equations --^ + t^ + ^ =1 and \ ~ jo + \ = 1 represent 
unparted hyperboloids, extending along the X and P-axes 
respectively; the hyperboloid in each case extending 
along the axis whose coordinate has the unique sign in 
the equation. 

When a = b, the equation becomes —+^5 — - = 1, 

a 1 a 1 (r 

which is the equation of an unparted hyperboloid of revo- 



238 



ANALYTIC GEOMETRY OF SPACE [Ch. V, § 30 



lution about the Z-axis. The equations — — + ^_ + * 



x 2 



* b 2 b 2 



and — — &- + — = 1 represent hyperboloids of revolution 
about the X and P"-axes respectively. 

30. The biparted hyperboloid. ^-f*-^=l.__ The 

a 1 b l c 2 
intercepts on the X-axis are ± a, but the surface does not 
cut the other axes. 





Eig. 18. 

The section of the surface made by the X-F-plane is an 
hyperbola, with its transverse axis 2 a along the X-axis ; 
the section made by the Z-X-plane is also an hyperbola, 
with its transverse axis 2 a along the X-axis ; the section 
by the P-^-plane is imaginary. 

Sections made by planes parallel to the Y-Z-plane are 
imaginary for values of x between + a and — a. When 
x = ± a, the sections are null ellipses, and for values of x 
numerically greater than a the sections are ellipses, in- 
creasing indefinitely as the cutting plane recedes from the 
origin. 

The sections of the surface made by planes parallel to 



Ch. V, § 30] 



QUADEIC SURFACES 



239 



the X-Y and Z-X planes are hyperbolas, and it may also 
be shown by the aid of transformation of coordinates that 
all planes through the X-axis are hyperbolas. 

This surface is called the hyperboloid of two sheets, or 
the biparted hyperboloid, extending along the X-axis. The 





Fig. 19. 



equations — ^- + ^_— ^=1 and — — — &- + — = 1 repre- 
a l o- c 2, a? o 2 cr 

sent biparted hyperboloids extending along the Y and 
Z-axes respectively. 

If = c, the equation becomes ~—^-z — - = 1, which is 

a 1 c 2 (r 

the equation of a biparted hyperboloid of revolution about 
the X-axis. The equations, 

--2"+M--2 = 1 ' and -m"-M + -9 =s1 ' 

or o z a A o- b- c- 

represent biparted hyperboloids of revolution about the 
I" and Z-axes respectively. 



240 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 31 

X u z 

31. The cone. -5 + *k — o = 0- — A cone, or conical 
a 1 ¥ c z 

surface, is a surface generated by a straight line passing 
through a fixed point, called the vertex, and always touch- 
ing some fixed curve. Any position of the generating 
line is called an element of the cone. 

When B = 0, we have seen (Art. 27) that the equation 

rp2i nii spZ 

of the second degree reduces to — ± f- ± — = 0. If both 

a 1 ¥ c 1 

the positive signs are used, the equation is satisfied by 
the coordinates of the origin only, and is therefore said 
to represent a null ellipsoid. If any other combination 
of signs is used, it will be shown to represent a cone. 

/yU niL iy2i 

Consider the equation — -f &- — - = 0. 
a 1 ¥ c 1 

The section made by the X-F-plane is a null ellipse; 
the sections made by the Y-Z and Z-X-planes are pairs 
of intersecting lines. 

Sections parallel to the X-JT-plane are ellipses, increas- 
ing indefinitely in size as the cutting plane recedes from 
the origin. Sections parallel to the other coordinate 
planes are hyperbolas. 

Moreover, if (x v y v z^) is a point on the surface, then 
any other point (kx v hy v Jcz^ on the line joining P x with 
the origin will also lie on the surface ; hence the surface 
is generated by a straight line passing through the origin, 
and is a cone extending along the Z-axis. 

The equations 

zyi2 npi g" !£& y2i git 

a 2 ¥ c 2 a 2 ¥ c 2 

or the same equations with their signs changed, represent 
cones extending along the X and Z-axes respectively, the 



Ch. V, § 32] QUADRIC SURFACES 241 

cone in each case extending along the axis whose coordi- 
nate has the unique sign in the equation. 

If the coefficients of the two terms which have the 
same sign are equal, the equation will represent a cone 
of revolution about the other axis. 

32. Asymptotic cones. — The equation of the imparted 
hyperboloid in polar coordinates is 

2 /cos 2 a cos 2 /3 _ cos 2 y\ _ -, 

p W 2 " b 2 ~~&~) ' 
1 

or p = . 

_!_ /cos 2 a cos 2 /3 cos 2 7 

There will, therefore, be real points on the surface for 
those values only of a, /3, and 7 which make 

cos 2 a cos 2 6 _ cos 2 7 p. 
a 2 b 2 c 2 

Let a', yS', and 7' be values of «, /3, and 7, for which 
this expression vanishes. Then, as «, /3, and 7 approach 
a', /3', and 7', the value of p will increase indefinitely, 
and the line through the origin whose direction angles 
are a\ /3', and 7' may be said to meet the surface at 
infinity. Such a line is called an asymptotic line. 

Since the equation 1 — — — — — = is the 

a 2 o 2 c 2 

only condition which must be satisfied by the polar coordi- 
nates of the points on all the asymptotic lines, it must 
be the equation of the asymptotic cone, which contains 
all these asymptotic lines of the surface. Multiplying 



242 



ANALYTIC GEOMETRY OF SPACE [Ch. V, § { 



by p' 2 and transforming to rectangular coordinates, it 
becomes 

^ + T _ t. = 

a 2 b 2 c 2 

Similarly it may be shown that the asymptotic cone 
of the biparted hyperboloid is 



:=0. 



33. The paraboloids. 



x 2 



b 2 



= 2 ez. — The surface 



= 2 cz passes through the origin, but does not cut 
the axes at any other point. Sections made by planes 





Fig. 20. 



Fig. 21. 



parallel to the X-Z"-plane are ellipses whose axes increase 
as the section recedes from the origin. Sections made by 
planes parallel to the other coordinate planes are parabo- 



Ch. V, § 33] 



QUADRIC SURFACES 



243 



las, which have their axes parallel to the Z-axis. This 
surface is shown in Fig. 20. It is called an elliptic parabo- 




Fig. 22 



loid. If b = a, — + -(- = 2 cz represents a paraboloid of 
a 1 a 1 

revolution about the Z-axis. 




Fig. 23. 



244 ANALYTIC GEOMETRY OF SPACE [Ch. V. § 33 

Let the student discuss the form of the surface repre- 
sented by the equation ^- — ^- = 2cz. It is called an 
hyperbolic paraboloid. (See Fig. 22.) 



PROBLEMS 

1. Prove that in both the elliptic and hyperbolic parabo- 
loids the sections parallel to the X-Z-plane are equal parab- 
olas ; also that the sections parallel to the F-Z-plane are equal 
parabolas. 

2. Show from the results of problem 1 that a paraboloid 
may be generated by the motion of a parabola, whose vertex 
moves along a parabola lying in a plane, to which the plane 
of the moving parabola is perpendicular ; the axes of the two 
parabolas being parallel, and (a) in the elliptic paraboloid, their 
concavities turned in the same direction ; (b) in the hyperbolic 
paraboloid, their concavities turned in opposite directions. 

3. Show that an ellipsoid may be generated by the motion 
of a variable ellipse, whose plane is always parallel to a fixed 
plane, and which changes its form in such a manner that the 
extremities of its axes lie in two ellipses, which have a com- 
mon axis, and whose planes are perpendicular to each other 
and to the plane of the moving ellipse. 

4. Find the equation of the cone, whose vertex is at the 
centre of an ellipsoid, and which passes through all the points 
of intersection of the ellipsoid and a given plane. 

5. Find the equation of the cone, whose vertex is at the 
centre of an ellipsoid, and which passes through all the points 
common to the ellipsoid and a concentric sphere. 

6. If a, b, c is the order of magnitude of the semi-axes of 
the ellipsoid in problem 5, and if the radius of the sphere 
is b, show that the cone breaks up into a pair of planes, 
whose intersections with the ellipsoid are circles. 



Ch. V, § 34] QUADRIC SURFACES 245 

34. Ruled surfaces. — A surface, through every point 
of which a straight line may be drawn so as to lie entirely 
in the surface, is called a ruled surface. Any one of these 
lines which lie on the surface is called a generating line 
of the surface. 

The cylinder and cone are familiar examples of such 
surfaces. We shall now show that the unparted hyper- 
boloid and the hyperbolic paraboloid are also ruled surfaces. 

The equation of the unparted hyperboloid may be 
written in the form 

X* z 2 y 2 

a 2 ~d* = W 



(H)S 



H'+i '"I 



If now we write the two equations 



z 



+-,=M 1 



a ' c '" 1 V" ' b. 



a c k x \ b 

in which k x may have any value, it appears that every 
point, whose coordinates simultaneously satisfy these 
equations, will satisfy the equation of the hyperboloid, 
and will therefore lie on the surface. But these two 
equations, used simultaneously, are the equations of a 
line, and, from what Ave have shown, that line must lie 
wholly in the surface. But since ^.may have any value, 
there will be an indefinite number of such lines, and it 
may be easily shown that one of them passes through 
each point of the surface. 



246 ANALYTIC GEOMETRY OF SPACE [Ch .V, § 34 

In the same manner it may be shown that there is 
another set of lines whose equations are 

*+? = *Yi-f\ *_? =i L(i + f\ 

a c J \ bj a c k 2 \ bj 

which lie wholly in the surface. A line of this set may 
also be passed through any point of the surface. Hence, 
through any point on this ruled surface, there may be 
passed two lines which lie wholly in the surface. Each 
line of one set cuts ever}^ line of the other set, but does 
not cut any line of the same set. 

Let the student show that the hyperbolic paraboloid is 
also a ruled surface. Figures 17 and 23 show the two 
sets of generating lines on both these surfaces. None 
of the other conicoids are ruled surfaces. 

PROBLEMS 

1. Prove that, if a plane is passed through a generating 
line of a conicoid, it will also cut it in another generating line. 
Will the two generating lines belong to the same set ? 

2. Prove that every generating line of the ruled paraboloid 

is parallel to one of the planes - ± - = 0. 

a b 

3. Obtain the equations of the generating lines which pass 
through the point (x lf y lt z^ of (a) the ruled paraboloid, 
(6) the ruled hyperboloid. 

4. Prove that the plane, which is determined by the centre 
and any generating line of a ruled hyperboloid, cuts the sur- 
face in a parallel generating line, and touches the asymptotic 
cone in an element. 

5. Show that, in both the ruled hyperboloid and the ruled 
paraboloid, the projections of the generating lines on the prin- 
cipal planes are tangent to the principal sections. 



Ch. V, § 35] QUADRIC SURFACES 247 

35. Tangent planes. — A tangent line to a surface may 
be denned as follows : Through P x and P v two adjacent 
points on the surface, draw a secant line. The limiting 
position, which this secant approaches as P 2 approaches 
P v is called a tangent line to the surface at the point P v 

Since P 2 may approach P x along the surface in an 
indefinite number of ways, there will be, in general, an 
indefinite number of tangent lines at any point of a sur- 
face. These will, in general, lie in a plane Avhich is called 
the tangent plane at the point P v 

We shall obtain the equation of the tangent plane at 
the point P 1 of the ellipsoid 

t. + it + t = 1 

a 2 T h 2 T 6 .2 

Transforming this equation to parallel axes with the 
origin at Pj (by [12]), and then to polar coordinates 
(by [6]), we have as the equation of the ellipsoid in polar 
coordinates (origin at P 2 ) 

2 /cos 2 it cos 2 ft cos 2 7 

V a 2 b 2 c 2 J 

For every set of values of «, ft, 7 in this equation there 
will correspond two values of p ; one value will always be 
zero, which agrees with the fact that the origin is a point 
on the surface ; the other value is 

! ;/ 1 cos ft [ z 1 cos 7 



cos 2 a cos 2 ft ( 



248 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 35 

which gives the distance from P 1 to any second point P 
of the ellipsoid, measured along the secant whose direc- 
tion angles are «, /3, 7. 

Now let P 2 approach P 1 along the ellipsoid ; then the 
secant line through P x and P 2 will approach as its limit- 
ing position a tangent line at P v whose direction angles 
we shall call «', /3', 7'. That is, as P 2 approaches P v 

t -, x* cos a , y, cos /3 , z, cos 7 
p approaches zero, and -J — h , + - 1 — o — - a P _ 

t x-. cos a f y. cos B' , z, cos 7' -rj 1 ,1 

proaches -*> — - — - + ifl ^ 4- -J — — '—. Hence, by the 
a 2 b 2 cr 

theory of limits, 

x 1 cos a' y x cos /3' z t cos 7' _ ^ 
« 2 P c 2 " ' 

If (p r , a\ /3', 7') are the polar coordinates of any point 
on any one of the tangent lines through P v this equation 
expresses the only relation which must hold between those 
coordinates, and is therefore the polar equation (referred 
to P 1 as origin) of the locus of the tangent lines through 
P v Multiplying by p' and transforming to rectangular 

coordinates (by [6]) we have ^ + ^f- + -\ = 0. Again 

a 2 b 2 c 2 

transforming to the original origin (by [12]), we have 

lf + ^+f = i> [so] 

as the required equation of the tangent plane. 

Let the student show that the equations of the tangent 
planes to the hyperboloids, 

6l 4- #?_ — — 1 
a* J* ^~ ' 

fiff ± WE.M Tlj [31] 



Ch. V, § 36] QUADKIC SURFACES 2<±9 

the paraboloids, — ± ^- = 2 cz, 

a 1 o 2, 

are *i^ ± m2L = c{z + Zi) . r 3 2] 

a 2 b 2 L -' 

36. Normals. — The line perpendicular to the tangent 
at the point of contact is called the normal to the surface 
at that point. 

Its equation for any particular surface can be easily 
obtained from the definition. 

PROBLEMS 

1. Prove that every tangent plane to a cone passes through 
the vertex. 

2. Prove that all the normal lines of a sphere pass through 
the centre of the sphere. 

3. Show that the length of a tangent to a sphere from the 
point (x l} y lt z x ) is the square root of the quantity obtained by 
substituting (a^, y 1} z x ) for (x, y, z) in the equation of the 
sphere. 

4. Show that the locus of points from which equal tangents 
may be drawn to a sphere is a plane. This plane is called the 
radical plane of the two spheres. 

5. Prove that the radical planes of three spheres meet in a 
line. This line is called the radical axis of the three spheres. 

6. Prove that the radical plane of two spheres is perpen- 
dicular to their line of centres. 

7. Prove that the radical axis of three spheres is perpen- 
dicular to the plane of their centres. 

8. Show (from its definition) that the tangent plane at a 
point Pj of a ruled surface contains the two generating lines 
of the surface which pass through P v 



250 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 37 

9. Prove that every plane which contains a generating line 
of a ruled surface is tangent to the surface at some point on 
the generating line. 

37. Diametral planes. — The locus of the middle points 
of a set of parallel chords of a quadric surface will be 
found to be a plane. This plane is called a diametral 
plane. 

Let a v ft v y 1 be the direction angles of a set of parallel 
chords in the ellipsoid, and let (V. y\ z') be the coordi- 
nates of the middle point P' of any one of these chords. 

Transform the equation of the ellipsoid to polar coor- 
dinates with P' as origin. Its equation (by [12] and 
[6]) is 

2 A? os 2 « cos 2 ft cos 2 7\ g fx'coa a y' cos ft z'cosy 

rJ2 0/ '2 «'2 
CI 2 b 2 C 2 

The two values of p given by this equation are the two 
distances from the origin to the ellipsoid, measured along 
a line whose direction angles are «, ft, y. If «, ft, y have 
the particular values a v ft v y v the two distances are 
equal but opposite in sign ; the sum of the roots of the 
equation, regarded as a quadratic in p, will be zero, and 
(by Introduction, Art. 8), 

x' cos a x y' cos ft r z' cos 7 1 _ ^ 
~<#~ b 2 c 2 ~ ' 

But x\ y', z' are the coordinates of any point on the 
required locus, referred to the original axes. Hence 

[33] 



Ch. V, § 37] QUADRIC SURFACES 251 

is the equation of the diametral plane bisecting the chords 
of the ellipsoid whose direction angles are « r fi v y v 

The locus is evidently a plane passing through the 
centre of the ellipsoid, and it is easily seen that any 
plane passing through the centre will be a diametral 
plane bisecting some system of parallel chords. 

Let the student show that the equations of the diametral 
planes of the hyperboloids, 

x 2 y2 £,2 QC cos ai y cos pi Z COS 71 _ Q . r->_n 

the paraboloids, 

^ ± £ = 2 «, are ^ ± l^ft = c fos Y , [35 ] 

From the last equation it appears that the diametral 
plane of a paraboloid is always parallel to the axis of 
the paraboloid. 

The line of intersection of any two diametral planes is 
called a diameter. All diameters of the central quadrics 
evidently pass through the centre ; in the paraboloids they 
are parallel to the axis. 

It may be shown that in the central quadrics there are 
three diameters which are so related that the plane of any 
two bisects all chords parallel to the third. Such diame- 
ters are said to be conjugate to each other ; and the plane 
through any two of them is conjugate to the third. 

PROBLEMS 

1. Obtain the equation of the diametral plane conjugate to 
a diameter through the point (x 1} y 1} z x ) of (a) the ellipsoid^ 
(b) the hyperboloids, (c) the paraboloids. (See [6].) 



252 ANALYTIC GEOMETRY OE SPACE [Ch. V, § 38 

2. Show that the tangent planes at the extremities of a 
diameter are parallel to the diametral plane conjugate to the 
given diameter. 

3. Show that the relation which exists between the direction 
cosines of any pair of conjugate diameters is 

cos «! cos a 2 cos fii cos fi 2 cos y! cos y 2 _ 
a 2 + 6 1 + ? ' 

4. Prove that the diametral plane of a sphere is perpendic- 
ular to the chords which it bisects, and that conjugate diame- 
ters are perpendicular to each other. 

5. Prove that every plane which passes through the centre 
of a central conic, or is parallel to the axis of a non-central 
conic, is a diametral plane, and find the direction cosines of the 
chords which it bisects. 

38. Polar Plane. — The locus of points which divide 
harmonically secants drawn from a fixed point to a quadric 
surface will be found to be a plane. It is called the polar 
plane of the given point with respect to the quadric sur- 
face. The fixed point is called the pole of the plane. 

We shall obtain the equation of the polar plane of the 
point P x with respect to the ellipsoid 

a 2 ^b 2 ^ <?~ 

We have seen that the polar coordinate equation of the 
ellipsoid, referred to P x as origin, is 

(cos 2 a cos 2 /3 cos 2 y\ fx 1 cosa j/jCOs/3 z 1 cosy 
^ a 2 ^ b 2 T <? 



Ch. V, § 38] QUADRIC SURFACES 253 

Through the origin P x pass a secant whose direction 
angles are «', /3', y'. Let the points where this secant cuts 
the ellipsoid be P 2 (p 2 , «', ft', 7') and P 3 (p 3 , «', 0', 7'), 
and on it locate a point P' (p', «', ft' , 7') sucli that 

1 " P^ + P.P,' H p 2 + Ps 
Then p 2 and p z are evidently the roots of the equation 

/cos 2 a' COS 2 /3' COh 2 7'\ 

p\-^-+-^-+^-J 

/ ^COStt' ^CO.S/3' Z : COS 7 '\ Xj 2 ,y! 2 Z, 2 _ 

+2p V~^ 2— +— ~^~ + ~- 2 ~ ) + ~^ + ~w +-?- 1 - 

Hence (by Introduction, Art. 8) 

^!_ h ^i 2 + !i 2 _l 

, a 2 b 2 c 2 



x 1 cosa' yjcosp 2^ cos 7 
— tf + P + ?~~ 



This is an equation connecting the polar coordinates of 
P', and is, therefore, the polar equation of the desired 
locus. Transforming to rectangular coordinates and to 
the original origin, we have, as the equation of the polar 
plane, 

2f+e« + « = l. [36] 

a 1 o- c 2 L J 

Let the student obtain the equation of the polar plane 
of a point with respect to each of the quadric surfaces. 



254 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 38 

PROBLEMS 

1. Prove that the polar plane of P, with respect to any 
quadric surface passes through the points of contact of all the 
tangent lines from P 1 to the surface. 

2. Prove that the polar planes of all points in a given plane 
pass through the pole of that plane 5 and, conversely, the poles 
of all planes passing through a given point lie on the polar 
plane of that point. 

3. Prove that the polar planes of all points on a given 
diameter of a quadric surface are parallel to the tangent plane 
at the extremity of the diameter. 

4. Prove that the polar plane of P x with respect to a sphere 
is perpendicular to the diameter through P v 

5. Prove that in the sphere the product of the distance of 
the pole from the centre and the distance of the polar plane 
from the centre is equal to the square of the radius. 

6. Prove that the distances of two points from the centre of 
a sphere are proportional to the distances of each from the 
polar plane of the other. 

LOCI PROBLEMS 

1. Find the locus of points which are equally distant from 
two intersecting planes. Show that it consists of two planes 
which are perpendicular to each other. 

2. Show that the locus of a point, the sum of the squares of 
whose distances from any number of points is constant, is a 
sphere. 

3. A point moves so that the sum of the squares of its dis- 
tances from the six faces of a cube is constant ; show that its 
locus is a sphere. 



Ch. V, § 38] QUADRIC SURFACES 255 

4. A and B are two fixed points, and P moves so that 
PA = nPB ; show that the locus of P is a sphere. Show also 
that all such spheres, for different values of n, have a common 
radical axis. 

5. Show that the locus of the point of intersection of three 
mutually perpendicular tangent planes to an ellipsoid is a circle 
about the centre of the ellipsoid, whose radius is Va 2 + b 2 -f c 2 . 

6. Show that the locus of the point of intersection of three 
mutually perpendicular tangent planes to a paraboloid is a 
plane. 

7. Find the locus of a point whose distance from a given 
point bears a constant ratio to its distance from a fixed plane. 

8. Three fixed points on a line lie, one in each coordinate 
plane ; find the locus of any fourth fixed point of the line. 

9. Show that the locus of the points, which divide in any 
given ratio all straight lines terminated by two fixed straight 
lines, is a plane. 

10. A line of constant length has its extremities on two 
fixed straight lines ; show that the locus of its middle point 
is an ellipse. 



ANSWERS. 



Page 10 

4 w g. I> (-!• !)• (-*• -!)■ (!• - 

W (|V2,o), (*JvS). (-|V2,o). (o, -|^)- 

5. (a) (a, 0), (6, c), (a + &, c). 
(6) (a, 0), (0, c), (a, c). 

6. (a) (0,0), (a,0), (|, fa). 

C) («-, o)._(-§. .0). (0, fa). 

CO (fva- o), (-|V3, 1). (-,Va, -f). 

Page 14 
2. \/34i VI30, 2\/29. 3. 5>/3, 2Vl3, V7. 



4. Va 2 + 6 2 , Va 2 + b 2 + ab\/2. 

Page 18 

2. (-19, -16). 12. |, --y-. 

3. (-11, 2). 13. (-9, 6). 

10. (11,5). 14. (a) (¥»-!)• (5) (1,-1). 

Page 23 

1. 6s-4y + 19 = 0. 6. 2 a- 2 + 2 i/ 2 + 14.x- 19 «/ +55 = 0. 

2. */ = 3a\ 7. x + y - 10 = 0. 

3. x 2 + y 2 + 6 x - 8 y = 0. 8. x 2 - 3 ?/ 2 = 0. 

4. 24x 2 + 25y 2 -250a* + 625 = 0. 9. 8x-2y + 17=0. 

5. x~ + y 2 - 5 a; + 5 y + 5 = 0. 10. a 2 + y 2 - x - y = 0. 

Page 32 
3. 2V5; fV2; |\/l70. 6. 6<^; 6>i; & = |. 

5. 6. 7. a- 2 -V 2 = 0. 

257 



258 ANSWERS 

Page 3S 

2. (a) 5x + 8*/ = 7. (c) x - 4 = 0. 
(6) 3x-4?/ = 0. (d) y - o = 0. 

3. x-3*/ = 3. 4. Yes. No. 

5. yi(x. 2 - Xi)+y 2 (x 3 - Xi) + y s (xi - x 2 ) = 0. 

6. 39 x - 79 y = 200. 

7. Equations of medians, a — ?/ — 1 = 0, 

x + 2 y + 1 = 0, 
x- 13^-9 = 0. 
Point of intersection, (i, — §). 

8. Equations of diagonals, bx + ay = ab, 

bx — ay = 0. 

Point of intersection. [^- - V 

Page 40 

1. VSx-y = 6(y/S + l). 

2. (a) V3 x - 3 y = - 18. (fr) 3 x - 5 «/ = 25. (c) x - y - - 3. 

3. 7V3x + 7y = llV3-2. 4. -2. 

Page 43 
2. (a) a = 10, & = -#, Z = |. 4. (a) x-8y + 5 = 0. 

(6) a = -|, 6 = |, 1= |. (6) 2x- y-2 = 0. 

(c) a = 0. 6 = 0, 7 = - 4. (c) 3 x + 5 y = 0. 

(d) a = — 4, b = co, Z = cc 

Page 46 

2. tan~ 1 (2if). 3. 135°. tan-!(7), tan-^Oi), tan-^lf). 

4. Exterior angle between first two lines. tan _1 (— ||). Opposite inte- 

rior angles, tan- 1 (|f), tan~ 1 (— 2 T 2 -). 

Page 47 

2. 7 >• — 3 ?/ = 11. 

3. («) ?/ = 0, 4 x + ^ - 24 = 0, 2 x - y = 0. 
(6) x-4?/ = 0, x + 2?/-6 = 0, x-4 = 0. 

(c) x - 3 = 0, x - 4 y + 11 = 0, x + 2 y - 10 = 0. 

(d) 4 x - 5 y = 0, x + y - 6 = 0, 8 x - y - 24 = 0. 

Page 49 

1. (6 + 5V3)x- 3y = 0. 2. 7x-3?/+5 = 0. 



ANSWERS 



259 



Page 51 

1. (a) x + V3 y - 10 = 0. (6) x - VS y + 10 = 0. 

(c) V3 x - 2/ + 10_= 0. (d) x + y = Q. 

(e) 4z+(2 + 2V'3_)?/-4v / 2 = 0. 
(/) 2x-(V2+V6>-12=0. 



1. T 3 3\/13. 
4. 4 « + 3 y ■ 



Page 54 
2. - 1. 

Page 57 
1. y + 7y+ 3 = 0; 7a; 



-.'/ 



3. 1H. 

6. T 2 T \ a/113 ; the first. 



17=0. 



Page 58 

2. 11 3 - 35 y = 0. 

3. 49 a + 98 y - 272 = ; 15 a; - 18 y - 320 = ; 4 x + 5 >/ - 3 = 0. 

4. 3 x + 2 y + 7 = 0. 5. \ 3 x-y + 3 — V& = 0. 6. 10 x - 3 y + 4 = 0. 



Page 60 



1. 32. 



3211. 



Page 60. General Problems 

2- to,#)i (*♦.#)■ 

3. 2 ;,• + 5 y = 40 ; 18 x + 5 ?/ = 120 ; 6(2 T V7> + 5(2 ± Vl)y =-. 120. 

4. 5 a; - y + 10 = 0. 5. 2 x + 3 y + 12 = ; C a; + // - 12 = 0. 

6. («) x-4i/ = 0. (/)) 2 .-• - 3 y - 10 = 0. 

7. (31, 5±-|V3). 8. (10, 5i). 

9. (16f, -0£); (4f, -M). 10- Mr, li° r ); *fr -V r ). 

n / 5\/lT)7 ± 3V82 4Vl97± 14V82 > 
V \/l97 ± V82 ' Vl97 ± V82 
(W2fl±3V82 ) -7V29J, 7V82 \ 
V29 ± V82 J 



V29 ± V82 
/ 4V29± 5V197 -9V29±2\/lTrp 



V V29 ± v'197 
5 x + y = 26. 



17 



V29 ± V197 



2(wi — m 2 ) 

Page 70 

2. 4 x - 3 ?/ + 15 = 0. 



18. 5. 



21. (5, 5). 



260 ANSWERS 

Page 71 

1. 3x+7?/-10 = 0. 5. (_4 ? 2). 

3. xy = 5- 6. | tan- 1 2. 

^ 4. tan-!f 

Page 74 

2. (a) p 2 = q2ft2 t . 4. («) rf + ?/ 2 - ay = o. 



a 2 cos 2 + & 2 sin 2 
2 m cos 



(5) (x - b) Vx 2 + y' 2 = as. 



• *>. ^ no cus (7 

( ' P = sin** ' ( c) * 2 +*/ 2 -«*-aVx 2 + 2/ 2 = 0. 



(d) (x 2 + ?/ 2 ) 2 = a 2 (x 2 - ? / 2 ). 



(e) p = a sin 2 6. 
(/) p = — a cos 0. 
2 a — cot 2 



W P cos 2d' (e) x 2 + 2/ 2 + &x-«vV 2 + ?/ 2 = 0. 

(a") /> 2 = a 2 cos 2 0. ^ Qtf + ^2)3 _ 4 ^2^2. 

(0) x 2 - ?/ 2 = a 2 . 

(A) (xH^^Cx 2 ^.^-*, 2 ) 2 . 

{g) P " cosfl " (0 (.^ 2 + 2/ 2 )^4a¥. 

(/*) p = a(l - 2 cos 0). ( j) x 3 + xy 2 -2 ay' 2 = 0. 

Page 78 

1. (a) x 2 + ?/ 2 + 4 x - 6 */ - 23 = 0. 
(6) x 2 + ?/ 2 + 6x + 8?/ = 0. 

(c) x 2 + j^-l0»-6y+ 3SJI = 0. 

(d) x 2 + y' 2 - 36 x - 32 y + 480 = 0, or 

x 2 + y' 2 - 4 x - 8 y - 80 = 0. 
(O 19 x 2 + 19 y 2 + 2 x - 47 y - 312 = 0. 
(/) 3 x 2 + 3 y 2 - 13 x - 11 y + 20 = 0. 
(fiO x 2 +?/ 2 -x-4?/-6 = 0. 
(h) 3 x 2 + 3 y 2 - 1 14a - 64 y + 276 = 0. 
(0 x 2 +?/ 2 -(5±Vif)x-(3±5VI|) ? /-18±30v / J|=0. 

2. (a) (-4, 3); V35. (c) (0, -3); 5. 
(6) Imaginary. (d) (£, 0); 1V145. 

6. 6 x + 3 ?/ - 10 = 0. 

Page 82 

1. (a) 3x + 40 = 25 ; 4x - 3y = 0. 
(5) Indeterminate. 

(c) 3 x + 7 y = 93 ; 7 x - 3 ?/ = 43. 
3 x — 7 y = 65 ; 7 x + 3 y = 55. 

(d) x + 5 y = 114 ; 5 x - 6 ?/ = - 27. 
6x-5y= 44; 5x + 6?/ = 57. 

2. 45°. 5= V47. 



ANSWERS 261 

Page 84 

1. (a) (21 ±4V51)x + (28 + 3 VST)*/ = 350. 

(b) x + 5y-2$=0; 5 x - y + 16 = 0. 

(c) 2 a; - y = 15 ; 58 x + 71 y = 335. 

2. (a) 6 x + 8 y - 49 = 0. (c) Ux + 3y = 55. 
(6) 2 £ - 3 y + 9 = 0, 

3. x x x + ?/i 2/ = r 2 . 

Page 85 

1. (a) 3 x - 2 y ± 7VI3 = 0. (6) 2 .r + 3y ± 7 V13 = 0. 

2. 3x + ?/ + 9±3VlO = 0. 3. 1 + -U=I. 

a' 2 & 2 r 2 

4. ^ = 36 ± 20V6. 

5 . „4 2 # 2 + &D 2 + 1BCE-2A BD E-A A 2 F - A C 2 + 4 J. CD - 4 D 2 ^ = 0. 



1. 

3. 
5. 


Page 86 

$ V21. 2. x 2 + ?/ 2 + 52 x - 21 ?/ - 265 

2 x 2 + 2 y 2 - 13 x - 6 y + 15 = 0. 

/ 64^20 + 375X2, / , 2V26 + -i35\ 2 _ / 501 \ 2 


^ 25V2(>+170^ ' \ ' 25V20 + 170^ V.25V26 + 170/ 1 


6. 


-^—- 7. 2* + y = 2; sc-2y = 0; (0.0), (f, |); tan 



2 /1//1 

Page 89 

3. Perpendicular bisector of the line joining the two points. 

5. Circle of radius r about (x i? y{). 

6. Perpendicular bisector of the line joining the two points. 

7. Bisector of the angle between the lines. 

8. Circle about the centre of the square. 

9. Circle whose centre is on the line through the fixed point, perpendic- 

ular to the fixed line. 

11. Circle whose centre is on the base of the triangle, extended. 

12. Circle whose centre is at the centre of the triangle. 

14. Circle whose centre is at the intersection of the two lines. 

15. Circle whose centre is on the line OX. 

18. Line through the centre of the base and the centre of the altitude of 

the triangle. 

19. A straight line. 

20. Two lines through the origin. 

21. A line through the origin. 

22. A circle. 

23. x' 2 + y 2 - rVx 2 + y 2 = ry. 



262 ANSWERS 

24. A diagonal of the rectangle. 

25. A diagonal of the parallelogram. 

30. x 2 + y 2 *ib_ fax + yyy) = ( ^^ ) r 2 . 

31. XiX + ijxy = r 2 . 

32. An equal circle tangent to the given circle at the fixed point. 

33. {;xi 2 ^y{ 2 -r 2 )[{x-x l y 2 +{y-y 1 ) 2 ]+2k 2 {x l x + y l y-x l 2 -y 1 2 )^^ = ^ 

34. A line parallel to the fixed line. 
35 A circle. 

Page 103 

1. (a) y 2 = — 2 mx. (6) x 2 = 2 my. (c) x 2 = — 2 wt«/. 

2. ?/' 2 = 2 mx + m 2 . 4. 4 x' 2 = — 9 y. 

3. (?/ + /3) 2 = 2 m (a- + a). 5. x = - 2 ; (2, 0) ; 4. 

Page 110 

,- ,- 9 

1. («) a = 3, b = 2, c=V5, e = ±v5, x = ±- L ^- 

V5 

(6) a = 3, b — 2, c = V5, e = iVo, ?/ = ± —3- 

(c) a = iV30, 6 = iV5, c = |V3, e = 1VI0, x-±fV3. 

2. (a) 4 x 2 + 9 y 2 = 36. (d) 16 x 2 + 25 ?/ 2 = 400. 
(&) 3 x 2 + 4 y 2 = 36. (e) 16 x 2 + 25 ?/ 2 = 400. 
(c) 5 a; 2 + 9 ?/ 2 = 180. (/) 8 x 2 + 9 ?/ 2 = 1152. 

5. 3 x 2 + 7 ?/ 2 = 55. 6. 

Page 115 

1. («) a = 5, b = 1, c = V26, e = aV26, 

(6) a = 2. & = 3, c = Vl3, e = |V13, 

(c) a=Vl0, 6=2, c=Vl4. e = iV35, x=±fVl4, 2x± VI6y=0. 

2. (a) 4 a: 2 - 9 y 2 = 36. (d) 16 x 2 - 9 ?/ 2 = 144. 
(6) 3 x 2 - y 2 = 9. (e) 9 x 2 - 16 y 2 = 144. 
(c) 5a 2 -4y 2 = 125. (/) 72 x 2 - 9^ = 800. 

3 (x-ct) 2 _(y-/3) 2 = 1 4 Impossible . 

a 2 b 2 

Page 118 

1. 4 x 2 _ |/2 _ _ 4 . a — i ? & — 2, e = § V5 ; latus rectum = 1 ; foci, 

(0, ±V5); directrices, y =± ^V5. 

2. V2. 4. 2 x?/ = a 2 . 



(a 


a' 2 


& 2 


2 - 


:1. 


X : 


V26 


x± 5 


y = 


0. 


c = 


V13 


3 x ± 


2y 


= 0. 



ANSWERS 263 

Page 123 
1. - X /V2; |V2. 2. |Vl3±i\/65. 

Page 130 

1. («) 3 x + 8 y = 19 ; 8 x - 3 y = 2. 
(6) 3 x + y = 7 ; x - 3 y = 9. 
(c) x ■+ 2 2/ = - ; 2x - y = 18. 
(V) 5 x - 6 «/ = - 8 ; 6 x + 5 y = 27. 
2- (a) ¥-; -I- (&) -I; 6. (c) -12; 3. 
(a) j/ = 4 ; 3 .>• + 2 y = 17. 4. (a) 12 x + 25 y = 100. 

(6) x - y = - 1 ; x + 3 y = - 9. {b) x + y = 3. 

('•) •'• + 3 y = 5 ; x - 3 y =- 7. (c) y = 3. 

(a) |V73; |V73. (6) |VlT); 2vT0. 
(c) 6\/6; 3 V5. 6 tan~i(± 3). 

Page 132 

x-2y±Vl7=0. 5 . /», ±m V 

18 x + 27 y = 88. \' 2 I 

_9±3Vn (±-^=. ±- 62 v 



r> ■'■" + ."- £ \ Va 2 +6 2 vV + /W 



4. /3 = ± Vft 2 - a 2 ? 2 . 



± - 



6 2 



Va 2 — b 2 Va? — b' 2 / 
Impossible in hyperbola when b > a. 



V v2 _ V'2/ V V2 vV 

8. 5y ± xVl§ ± 4\/l0 = 0. Four tangents. 

9. 7 y ± 2 xa/35 ± 4 a/91 = 0. Four tangents. 

10. x ±yV3 + Q = 0. Two tangents. 12. (a) 4. (6) 3. 

Page 148 

1. 3 x - 8 y = ; x - 3 y = 0. 5. 26 x + 33 y 

2. y + 9 = 0. 6. 8 x -f 45 y = ; 

3. 2 xV3 + 3 y = 0. 7. -^ V3. 

4. x + 2y = 8. 8 x-y = 1. 
9. (±iVl5, T |Vl5); (±|Vl5, T & Vl5). 

Page 158 

1. (a) x-Sy= 16. 
(6) x + 2y=-6. 



1ZD. 

.( 


45 8 


^ 


V ± Vl51 ±Vl5 


1/ 




10. ?i = - 


-h. 


00 


15x+ 16?/=- 


24. 


id) 


x + 5 = 0. 





264 

6- (-ff, ft)- 



ANSWERS 

4- (-10,4). «.(--"»). 

7 / « 2 fr 2 xi « 2 6 2 yi \ 

U 2 xr + a' 2 ?/i 2 ' 6 2 xi 2 + aVJ 



1. Imaginary ellipse. 

2. Real ellipse. 

3. Two intersecting lines. 

4. Hyperbola. 

5. Two parallel lines. 



2. The directrix. 



Page 182 

6. Parabola. 

7. Two coincident lines. 

8. Point, 

9. Point. 

10. Hyperbola. 

Page 186 

4. x=™-. 



5. (a) x 2 + y 2 = a 2 ; (6) x 2 + y 2 = a 2 ; (c) x = 0. 



+ ^ = 2 - 



a 2 Z> 2 



7. The asymptotes. 
10. & 2 x 2 + a°-y 2 = b 2 c 2 . 

12. l/2 = -^-X. 



a 4 6 4 a 2 b 2 



11. (a) An ellipse ; (b) A parabola. 
14. 25 x 2 + 16 w 2 - 48 y - 64 = 0. 



15. 2 r Vx 2 + y 2 - 2 xix - 2 y^ + Xi 2 + y{ 2 - r 2 = 0. 

16. x 2 -*/ 2 -fx + f y-| = 0. 

20. 3 x 2 - y 2 - 2 ex = 0. (Take the origin at the vertex of the smaller 
angle. ) 

23. x 2 + y 2 = 



22. A parabola. 

24. 2 xy - y x x - x^y = 0. 

27. y 2 = — 2 mx. 

30. x = ■ 



9 
( X 2 + ^2 _ 2 aiC )2 _ a 2 ( a _ x) 2 + a' 2 //' 2 . 



m + mi 



32. (x 2 + ?/ 2 ) 2 = a' 2 x 2 + b 2 y 2 . 
34. 4 &i 2 x 2 ?/ 2 - 4 «i 2 */ 4 = ai 2 &i 4 . 
36. A directrix. 

38. 



28. ab 2 Vx 2 + y2 = 62x2 + a 2 y 2 m 

31. X 2 + y*-(?^y+C* = 0. 

33 «i% 2 + &iV = :L 

a 4 6 4 



5. Circle with radius a + b. 



c2 [aW + b 2 x 2 ^ x 2 i£ = 1 
61 LaV 2 + & 4 x 2 J a 2 & 2 



x 2 y z -, 

— + ^7 = 1. 

a 2 q? 

b 2 



1. VW; (6, -10. 20)- / ^i + aa + a* .Vi + H* + ^ ^it^: 



ANSWERS 265 

Page 199 

3 ' 3 

Page 202 

1. a = /3 = 7 = cos" 1 1 Vo. 

13 1 

3. cos rt = — — , cos /3 = — — , cos 7 = — — ■ 

\U \ li Vn 

1 — 4 7 

4. cos a = . , cos /3 = — =r , cos 7 = — — 

\ lie, V06 V66 

1 2 3 

6. cos « = — — , cos /S = , cos 7 = — — « 

Vl4 VII Vl4 

8. 60°. 

Page 205 

2. cos-if. 4. (|V3, I, 0); (|V3. 0. |). 

5. (7, cos -1 -, cos -1 ^-^. cos -1 -V 6- cos -1 4 Via. 
^ 7 7 i) 

Page 208 

3. >j- - *■- = 2-3 : .v- + 2 2 = 25. 

4. x- + y 2 + 2 2 - .*• - 4 y - 10 2 + 13 = 0. 

5. 7 x + 7y+l0z = 9; 2x-y-7z = - 30. 

Page 211 

2. x 2 + 5- 2 - 6 .r - 10 .: + 9 = 0. 

3. (a) y 2 + z 2 = 16. (c) x s + y- + 2" 2 = r 8 . 
(6) jc 2 - >/ 2 - z- = 0. ((7) ^ + z 2 _ 2 )„j-. 

4. &-V- - aV - afts 3 = a 2 © 2 : We 2 - a-y- + fr 2 2 = a-b-. 

Page 218 

1. (a) * -=1. 4=- (6) -^r (O -^r 0?) 4, - f. 12. 

V74 v 74 \ 74 \ 74 V74 

2. $x + y-bz = 35. 6. 3 a; + 59 y - 72 z = 0. 

Page 221 

1. (a) 11 a; - 17 y - 13 s + 3 = 0. (©) 4s - 7 y - 5s - 8 = 0. 

2. 12 as - 17 y -f4z-3 = 0. 5. 2 x+z- 8 = 0. 

3. y + 3s + 3 = 0. e. 10a; - 19 y - 32 z = 0. 

4. 7s-f4y-4z-22 = G. 



266 ANSWERS 

Page 224 

1. x = y — ; y = z — ; z — x = 0. 

2. z = 0, Ax +By + C = 0, etc. ; z = k, Ax + By + C = 0, etc . 

z = ki, y = &o, etc. 

4. (a) is = f s + I, ^ = £0- f- («") 2/ = 2, = - 3. 
(6) x = - -^ 0, ?/ = - 2 ^ + 4. (e) x = | ?/ - |, = 1 
(c) = - 4, x = - 2 y + 3. (/) y = - 1, = 2. 

5. = 0, x = | // + 3 ; x = 0, = 3 y +'6 ; y = 0, = — 2 x + 6. 

Page 226 

2. (a) x = 2/ - 3, x = i V2 + |V2 - 1. 
(6) x = - y + 1, a: = iv'2 + f V2 - 1. 
(c) x = V3 y - 2V3 - 1, = - 3. 

3. x = y = z. 

Page 228 

3. • =£ ?• (0 » =L 0. 

it* v5 V5 

(^ -, -, -. (d) 1, 0, 0. 

v ' 3 3 3 V y ' 

Page 229 

3. (a) x = - 3 y, (b) x = \z-i (c) x = 2, 

= - 2 y - 2. y = - f + 8. y = - 3. 

4. x = 0, 2 y + = 0. 8. (4. 1, - 2). 9. 12 x - 5 y - - 31 = 0. 

Page 231 

1. (a) x 2 + y' J + 2 - 10 x + 4 y - G + 37 = 0. 
(6) x' 2 + y 2 + 2 - 4 x + 6 y + 12 = 0. 

(c) x 2 + y 2 + 2 ± 2 ax = 0. 

2. (a) (1, - 3, 4), 2. (a") (- 3, 0, 0), 3. 
(6) (—5, 2, 1), 5. 0) Imaginary, 
(c) (-2, -2, -3), 4. 

3. («) Indeterminate ; points lie in a plane. 
(6) x 2 + y 2 + 2 - 2 x - 8 y - 16 = 0. 

4. x 2 + y 2 + 2 - 2 x - 2 y - 2 = 0. 

5. x 2 4- 2/ 2 + 0' 2 = 4. 



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